Number Play Class 8 Mathematics NCERT Solutions

Q1Figure it Out 1

The sum of four consecutive numbers is 34. What are these numbers?

Solution

Given: The sum of four consecutive numbers is 34.

To Find: The four consecutive numbers.

Let: The four consecutive numbers be

x

,

x+1

,

x+2

, and

x+3

.

Equation: According to the question, their sum is 34.

x + (x+1) + (x+2) + (x+3) = 34

Solution:

4x + 6 = 34

4x = 34 - 6

4x = 28

x = \frac{28}{4}

x = 7

The numbers are: First number =

x = 7

Second number =

x+1 = 8

Third number =

x+2 = 9

Fourth number =

x+3 = 10

Verification:

7 + 8 + 9 + 10 = 34

Final Answer: The four consecutive numbers are 7, 8, 9, and 10.

Q2Figure it Out 1

Suppose $p$ is the greatest of five consecutive numbers. Describe the other four numbers in terms of $p$ .

Solution

Given:

p

is the greatest of five consecutive numbers.

To Find: The other four numbers in terms of

p

.

Solution: Since the numbers are consecutive, each number is 1 less than the next number. If the greatest number is

p

, the number before it is

p-1

. The number before

p-1

is

(p-1)-1 = p-2

. The number before

p-2

is

(p-2)-1 = p-3

. The number before

p-3

is

(p-3)-1 = p-4

.

So, the five consecutive numbers in descending order are

p, p-1, p-2, p-3, p-4

.

Final Answer: The other four numbers are

p-1

,

p-2

,

p-3

, and

p-4

.

Q3Figure it Out 1

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(i)

The sum of two even numbers is a multiple of 3.

(ii)

If a number is not divisible by 18, then it is also not divisible by 9.

(iii)

If two numbers are not divisible by 6, then their sum is not divisible by 6.

(iv)

The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

(v)

The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Solution

(i)

The sum of two even numbers is a multiple of 3. Answer: Sometimes True. Explanation: Let the two even numbers be

2m

and

2n

. Their sum is

2m + 2n = 2(m+n)

. This expression is not always a multiple of 3.

Example (True): Let the numbers be 2 and 4. Their sum is $2+4=6$ , which is a multiple of 3.
Non-example (False): Let the numbers be 4 and 8. Their sum is $4+8=12$ , which is a multiple of 3. Let's try another. Let the numbers be 2 and 8. Their sum is $2+8=10$ , which is not a multiple of 3.

(ii)

If a number is not divisible by 18, then it is also not divisible by 9. Answer: Sometimes True. Explanation:

Example (True): The number 10 is not divisible by 18, and it is also not divisible by 9.
Non-example (False): The number 27 is not divisible by 18, but it is divisible by 9.

(iii)

If two numbers are not divisible by 6, then their sum is not divisible by 6. Answer: Sometimes True. Explanation:

Example (True): The numbers 2 and 3 are not divisible by 6. Their sum is $2+3=5$ , which is not divisible by 6.
Non-example (False): The numbers 2 and 4 are not divisible by 6. Their sum is $2+4=6$ , which is divisible by 6.

(iv)

The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. Answer: Always True. Explanation: Let the multiple of 6 be

6k

and the multiple of 9 be

9m

, where

k

and

m

are integers. Their sum is

S = 6k + 9m

. We can factor out 3 from the expression:

S = 3(2k) + 3(3m) = 3(2k + 3m)

. Since

(2k + 3m)

is an integer, the sum

S

is always a multiple of 3.

(v)

The sum of a multiple of 6 and a multiple of 3 is a multiple of 9. Answer: Sometimes True. Explanation: Let the multiple of 6 be

6k

and the multiple of 3 be

3m

. Their sum is

S = 6k + 3m = 3(2k + m)

. For this sum to be a multiple of 9, the term

(2k+m)

must be a multiple of 3.

Example (True): Let $k=1, m=1$ . The numbers are 6 and 3. Their sum is $6+3=9$ , which is a multiple of 9.
Non-example (False): Let $k=2, m=1$ . The numbers are 12 and 3. Their sum is $12+3=15$ , which is not a multiple of 9.

Q4Figure it Out 1

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Solution

Given: A number

N

leaves a remainder of 2 when divided by 3. This can be written as

N = 3k + 2

for some integer

k

. The same number

N

leaves a remainder of 2 when divided by 4. This can be written as

N = 4m + 2

for some integer

m

.

To Find:

A few such numbers.
An algebraic expression for all such numbers.

Solution: From the given conditions, we can write:

N - 2 = 3k

N - 2 = 4m

This means that the number

(N-2)

is a common multiple of 3 and 4. The least common multiple (LCM) of 3 and 4 is 12. So,

(N-2)

must be a multiple of 12. We can write this as

N-2 = 12j

for some integer

j

. Therefore, the algebraic expression for all such numbers is

N = 12j + 2

.

To find a few such numbers, we can substitute different integer values for

j

: If

j=0

,

N = 12(0) + 2 = 2

. If

j=1

,

N = 12(1) + 2 = 14

. If

j=2

,

N = 12(2) + 2 = 26

. If

j=3

,

N = 12(3) + 2 = 38

.

Final Answer: A few such numbers are 2, 14, 26, 38. The algebraic expression to describe all such numbers is

12j + 2

, where

j

is a non-negative integer.

Q5Figure it Out 1

"I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up - it simply won't do, A stubborn odd pebble remains in my view. Group them by 5, yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?"

Solution

Given: A riddle about the number of pebbles, let's call it

N

.

When grouped by 3, remainder is 1: $N \equiv 1 \pmod{3}$ .
When grouped by 2 (pairing), remainder is 1: $N \equiv 1 \pmod{2}$ .
When grouped by 5, remainder is 1: $N \equiv 1 \pmod{5}$ .
When grouped by 7, remainder is 0 (perfection is found): $N \equiv 0 \pmod{7}$ .
The number is less than 100: $N < 100$ .

To Find: The number of pebbles,

N

.

Solution: From the first three conditions, we know that if we subtract 1 from

N

, the result is divisible by 3, 2, and 5. So,

(N-1)

is a common multiple of 2, 3, and 5. The least common multiple (LCM) of 2, 3, and 5 is

2 \times 3 \times 5 = 30

. Therefore,

(N-1)

must be a multiple of 30.

Possible values for

(N-1)

that are less than 100 are:

N-1 = 30 \implies N = 31

N-1 = 60 \implies N = 61

N-1 = 90 \implies N = 91

Now, we use the fourth condition: the number

N

must be divisible by 7. Let's check our possible values for

N

:

Is 31 divisible by 7? No, $31 = 7 \times 4 + 3$ .
Is 61 divisible by 7? No, $61 = 7 \times 8 + 5$ .
Is 91 divisible by 7? Yes, $91 = 7 \times 13$ .

This value also satisfies the condition

N < 100

.

Final Answer: The number of pebbles is 91.

Q6Figure it Out 1

Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tathagat's claim true?

Solution

Given: Numbers that leave a remainder of 2 when divided by 6 can be written in the form

6k+2

, where

k

is an integer. Tathagat's claim: The sum of any three such numbers is always a multiple of 6.

To determine: If the claim is true.

Let: The three numbers be

N_1

,

N_2

, and

N_3

.

N_1 = 6k_1 + 2

N_2 = 6k_2 + 2

N_3 = 6k_3 + 2

where

k_1, k_2, k_3

are integers.

Solution: Let's find the sum

S

of these three numbers.

S = N_1 + N_2 + N_3

S = (6k_1 + 2) + (6k_2 + 2) + (6k_3 + 2)

S = 6k_1 + 6k_2 + 6k_3 + 2 + 2 + 2

S = 6(k_1 + k_2 + k_3) + 6

S = 6(k_1 + k_2 + k_3 + 1)

Since

k_1, k_2, k_3

are integers, their sum plus 1, i.e.,

(k_1 + k_2 + k_3 + 1)

, is also an integer. Therefore, the sum

S

is of the form

6 \times (\text{an integer})

, which means it is always a multiple of 6.

Final Answer: Yes, Tathagat's claim is true.

Q7Figure it Out 1

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.

(i)

$4779+661$

(ii)

4779-661

Solution

Given:

661

divided by 7 gives a remainder of 3. So,

661 = 7k + 3

for some integer

k

.

4779

divided by 7 gives a remainder of 5. So,

4779 = 7m + 5

for some integer

m

.

(i) $4779+661$

Algebraic Solution: Sum =

(7m + 5) + (7k + 3)

=

7m + 7k + 5 + 3

=

7(m + k) + 8

Since 8 can be written as

7+1

, we have: =

7(m + k) + 7 + 1

=

7(m + k + 1) + 1

This expression is in the form

7 \times (\text{an integer}) + 1

. Thus, the remainder is 1.

Visual Explanation: Imagine the numbers as groups of 7s with some leftovers (remainders).

4779

is many groups of 7 and 5 leftovers.

661

is many groups of 7 and 3 leftovers. When we add them, we combine all the full groups of 7, and we also combine the leftovers:

5 + 3 = 8

leftovers. From these 8 leftovers, we can make one more full group of 7, with 1 leftover. So, the total sum has a remainder of 1.

Final Answer for (i): The remainder will be 1.

(ii) 4779-661

Algebraic Solution: Difference =

(7m + 5) - (7k + 3)

=

7m - 7k + 5 - 3

=

7(m - k) + 2

This expression is in the form

7 \times (\text{an integer}) + 2

. Thus, the remainder is 2.

Visual Explanation: We start with the groups of 7s and 5 leftovers from 4779. We need to subtract the groups of 7s and 3 leftovers from 661. We can subtract the full groups of 7 from other full groups of 7. Then we subtract the leftovers:

5 - 3 = 2

leftovers. So, the final result has a remainder of 2.

Final Answer for (ii): The remainder will be 2.

Q8Figure it Out 1

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Solution

Given: A number

N

such that:

$N$ leaves a remainder of 2 when divided by 3. ( $N \equiv 2 \pmod{3}$ )
$N$ leaves a remainder of 3 when divided by 4. ( $N \equiv 3 \pmod{4}$ )
$N$ leaves a remainder of 4 when divided by 5. ( $N \equiv 4 \pmod{5}$ )

To Find: The smallest such number.

Solution: Let's analyze the conditions:

$N = 3k + 2$ . This is equivalent to $N+1 = 3k+3 = 3(k+1)$ . So, $N+1$ is a multiple of 3.
$N = 4m + 3$ . This is equivalent to $N+1 = 4m+4 = 4(m+1)$ . So, $N+1$ is a multiple of 4.
$N = 5j + 4$ . This is equivalent to $N+1 = 5j+5 = 5(j+1)$ . So, $N+1$ is a multiple of 5.

From this, we can see that the number

(N+1)

is a common multiple of 3, 4, and 5. To find the smallest such positive number

N

, we need to find the smallest positive common multiple of 3, 4, and 5. This is the least common multiple (LCM) of 3, 4, and 5. Since 3, 4, and 5 are coprime, their LCM is their product. LCM(3, 4, 5) =

3 \times 4 \times 5 = 60

.

So, the smallest positive value for

(N+1)

is 60.

N+1 = 60

N = 60 - 1

N = 59

Explanation of why it is the smallest: We are looking for the smallest positive integer

N

that satisfies the conditions. Our method found that

(N+1)

must be a common multiple of 3, 4, and 5. The set of positive common multiples is {60, 120, 180, ...}. To get the smallest positive

N

, we must choose the smallest positive common multiple for

(N+1)

, which is the LCM, 60. Any other common multiple would result in a larger value for

N

.

Final Answer: The smallest such number is 59.

Q1Figure it Out 2

Find, without dividing, whether the following numbers are divisible by 9.

(i)

123

(ii)

405

(iii)

8888

(iv)

93547

(v)

358095

Solution

Rule for Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) 123 Sum of digits =

1 + 2 + 3 = 6

. Since 6 is not divisible by 9, the number 123 is not divisible by 9.

(ii) 405 Sum of digits =

4 + 0 + 5 = 9

. Since 9 is divisible by 9, the number 405 is divisible by 9.

(iii) 8888 Sum of digits =

8 + 8 + 8 + 8 = 32

. Since 32 is not divisible by 9, the number 8888 is not divisible by 9.

(iv) 93547 Sum of digits =

9 + 3 + 5 + 4 + 7 = 28

. Since 28 is not divisible by 9, the number 93547 is not divisible by 9.

(v) 358095 Sum of digits =

3 + 5 + 8 + 0 + 9 + 5 = 30

. Since 30 is not divisible by 9, the number 358095 is not divisible by 9.

Q2Figure it Out 2

Find the smallest multiple of 9 with no odd digits.

Solution

Given: We need to find a number that is a multiple of 9. The number must not contain any odd digits. The allowed digits are {0, 2, 4, 6, 8}. The number must be the smallest possible.

To Find: The smallest such number.

Solution: For a number to be a multiple of 9, the sum of its digits must be a multiple of 9. To make the number as small as possible, we should use the fewest possible digits and use smaller digits in the higher place values.

Let's find the smallest possible sum of digits from the allowed set {0, 2, 4, 6, 8} that is a multiple of 9. The smallest non-zero multiple of 9 is 9. Can we form a sum of 9 using only even digits? No, because the sum of any number of even digits is always even, and 9 is odd. The next multiple of 9 is 18. Can we form a sum of 18 using even digits? Yes.

To find the smallest number, we want the fewest digits.

Can we use two digits? The maximum sum of two even digits is $8+8=16$ , which is not 18.
We must use at least three digits. To make the three-digit number smallest, the first digit should be the smallest possible non-zero even digit, which is 2. If the first digit is 2, the remaining two digits must sum to $18 - 2 = 16$ . The only way to get a sum of 16 from two even digits is $8+8$ . So, the digits are 2, 8, 8. The smallest number we can form with these digits is 288.

Let's check if we can form a smaller number using more digits. A four-digit number would start with 2000... and would be larger than 288. So, the three-digit number 288 is the smallest.

Verification: The number is 288. It has no odd digits. The sum of its digits is

2+8+8=18

, which is a multiple of 9. So, 288 is a multiple of 9.

Final Answer: The smallest multiple of 9 with no odd digits is 288.

Q3Figure it Out 2

Find the multiple of 9 that is closest to the number 6000.

Solution

Given: The number 6000.

To Find: The multiple of 9 closest to 6000.

Solution: First, we divide 6000 by 9 to see what the remainder is.

6000 \div 9

6000 = 9 \times 666 + 6

This tells us that 6000 is 6 more than a multiple of 9.

The multiple of 9 just below 6000 is

6000 - 6 = 5994

. The distance between 6000 and 5994 is 6.

The multiple of 9 just above 6000 is

5994 + 9 = 6003

. The distance between 6000 and 6003 is 3.

Comparing the distances, 3 is smaller than 6. Therefore, 6003 is closer to 6000 than 5994 is.

Final Answer: The multiple of 9 closest to 6000 is 6003.

Q4Figure it Out 2

How many multiples of 9 are there between the numbers 4300 and 4400?

Solution

Given: The range of numbers is between 4300 and 4400.

To Find: The count of multiples of 9 in this range.

Solution: Step 1: Find the first multiple of 9 after 4300. Divide 4300 by 9.

4300 = 9 \times 477 + 7

. The multiple of 9 before 4300 is

9 \times 477 = 4293

. The first multiple of 9 after 4300 is

4293 + 9 = 4302

.

Step 2: Find the last multiple of 9 before 4400. Divide 4400 by 9.

4400 = 9 \times 488 + 8

. The last multiple of 9 before 4400 is

4400 - 8 = 4392

.

Step 3: Count the number of multiples. The multiples of 9 form an arithmetic sequence: 4302, 4311, ..., 4392. Let the first term be

a_1 = 4302

, the last term be

a_n = 4392

, and the common difference be

d=9

. Using the formula for the nth term of an arithmetic sequence:

a_n = a_1 + (n-1)d

.

4392 = 4302 + (n-1)9

4392 - 4302 = (n-1)9

90 = (n-1)9

10 = n-1

n = 11

There are 11 multiples of 9 between 4300 and 4400.

Final Answer: There are 11 multiples of 9 between 4300 and 4400.

Q1Figure it Out 3

The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

Solution

Given: Let the 8-digit number be

N

. The digital root of

N

is 5.

To Find: The digital root of the number

N+10

.

Concept: The digital root of a number is the single-digit value obtained by an iterative process of summing digits, on a given base. It is also the remainder when the number is divided by 9 (with the exception that a remainder of 0 corresponds to a digital root of 9). So, a digital root of 5 means

N \equiv 5 \pmod{9}

.

Solution: We want to find the digital root of

N+10

. This is equivalent to finding the remainder of

(N+10)

when divided by 9. We know

N \equiv 5 \pmod{9}

. And

10 \equiv 1 \pmod{9}

. Therefore,

N+10 \equiv 5+1 \pmod{9}

.

N+10 \equiv 6 \pmod{9}

. The remainder is 6. So, the digital root is 6.

Final Answer: The digital root of 10 more than that number will be 6.

Q2Figure it Out 3

Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Solution

Let's start with a number, for example, 7.

Sequence of numbers:

7

7 + 11 = 18

18 + 11 = 29

29 + 11 = 40

40 + 11 = 51

51 + 11 = 62

62 + 11 = 73

73 + 11 = 84

84 + 11 = 95

95 + 11 = 106

106 + 11 = 117

Digital roots of the sequence: Digital root of 7 is 7. Digital root of 18 is

1+8=9

. Digital root of 29 is

2+9=11 \rightarrow 1+1=2

. Digital root of 40 is

4+0=4

. Digital root of 51 is

5+1=6

. Digital root of 62 is

6+2=8

. Digital root of 73 is

7+3=10 \rightarrow 1+0=1

. Digital root of 84 is

8+4=12 \rightarrow 1+2=3

. Digital root of 95 is

9+5=14 \rightarrow 1+4=5

. Digital root of 106 is

1+0+6=7

. Digital root of 117 is

1+1+7=9

.

The sequence of digital roots is: 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, 9, ...

Observations: The sequence of digital roots forms a repeating pattern. The pattern is: 7, 9, 2, 4, 6, 8, 1, 3, 5. This pattern has a length of 9 and will repeat continuously.

Explanation: The digital root of 11 is

1+1=2

. Each time we add 11 to a number, we are effectively adding 2 to its digital root (modulo 9). For example, the digital root of 7 is 7. The next digital root is the digital root of

(7+2) = 9

. The next is the digital root of

(9+2) = 11 \rightarrow 2

. The next is the digital root of

(2+2) = 4

, and so on. This explains the repeating cycle.

Q3Figure it Out 3

What will be the digital root of the number $9a+36b+13$ ?

Solution

Given: The expression

9a+36b+13

, where

a

and

b

are integers.

To Find: The digital root of the value of this expression.

Concept: The digital root of a number is related to its remainder when divided by 9. We can find the digital root of a sum by finding the digital root of the sum of the digital roots of its terms.

Solution: Let's analyze each term:

$9a$ : Since $a$ is an integer, $9a$ is always a multiple of 9. The digital root of any multiple of 9 is 9.
$36b$ : Since $b$ is an integer and $36 = 9 \times 4$ , $36b$ is always a multiple of 9. The digital root of any multiple of 9 is 9.
$13$ : The digital root of 13 is $1+3=4$ .

Now, we find the digital root of the sum of the digital roots: Digital root of (9 + 9 + 4) = Digital root of (22). Digital root of 22 is

2+2=4

.

Alternatively, using modulo arithmetic: We need to find

(9a+36b+13) \pmod{9}

.

9a \equiv 0 \pmod{9}

.

36b \equiv 0 \pmod{9}

.

13 \equiv 4 \pmod{9}

. So,

(9a+36b+13) \equiv (0+0+4) \pmod{9} \equiv 4 \pmod{9}

. The remainder is 4, so the digital root is 4.

Final Answer: The digital root of the number

9a+36b+13

will always be 4.

Q4Figure it Out 3

Make conjectures by examining if there are any patterns or relations between

(i)

the parity of a number and its digital root.

(ii)

the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Solution

(i) The parity of a number and its digital root.

Examination:

Even numbers:
- DR(2) = 2 (Even)
- DR(10) = 1 (Odd)
- DR(12) = 3 (Odd)
- DR(20) = 2 (Even)
Odd numbers:
- DR(3) = 3 (Odd)
- DR(11) = 2 (Even)
- DR(13) = 4 (Even)
- DR(21) = 3 (Odd)

Conjecture: There is no direct or consistent relationship between the parity (even or odd) of a number and the parity of its digital root. An even number can have an even or an odd digital root, and an odd number can also have an even or an odd digital root.

(ii) The digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Relation with remainder when divided by 9:

Examination:

Number 17: $17 \div 9$ gives remainder 8. DR(17) = $1+7=8$ .
Number 48: $48 \div 9$ gives remainder 3. DR(48) = $4+8=12 \rightarrow 1+2=3$ .
Number 81: $81 \div 9$ gives remainder 0. DR(81) = $8+1=9$ .

Conjecture: The digital root of a number is the same as the remainder when that number is divided by 9. If the remainder is 0, the digital root is 9. Otherwise, the digital root is the remainder itself.

Relation with remainder when divided by 3:

Examination:

Number 17: $17 \div 3$ gives remainder 2. DR(17) = 8. $8 \div 3$ gives remainder 2.
Number 48: $48 \div 3$ gives remainder 0. DR(48) = 3. $3 \div 3$ gives remainder 0.
Number 25: $25 \div 3$ gives remainder 1. DR(25) = 7. $7 \div 3$ gives remainder 1.

Conjecture: The remainder obtained when a number is divided by 3 is the same as the remainder obtained when its digital root is divided by 3. (If the digital root is 3, 6, or 9, the remainder is 0. If it is 1, 4, or 7, the remainder is 1. If it is 2, 5, or 8, the remainder is 2).

Q1Figure it Out 4

If $31z5$ is a multiple of 9, where $z$ is a digit, what is the value of $z$ ? Explain why there are two answers to this problem.

Solution

Given: The number

31z5

is a multiple of 9.

To Find: The value(s) of the digit

z

.

Solution: According to the divisibility rule of 9, a number is divisible by 9 if the sum of its digits is a multiple of 9. Sum of the digits of

31z5

is

3 + 1 + z + 5 = 9 + z

.

For the number to be a multiple of 9, the sum

(9+z)

must be a multiple of 9. Since

z

is a single digit, its value can range from 0 to 9.

Let's check the possible values for

(9+z)

:

If $z=0$ , the sum is $9+0=9$ . Since 9 is a multiple of 9, $z=0$ is a valid solution.
If $z=9$ , the sum is $9+9=18$ . Since 18 is a multiple of 9, $z=9$ is a valid solution.
For any other digit $z$ (from 1 to 8), the sum $(9+z)$ will be between 10 and 17, none of which are multiples of 9.

Explanation for two answers: There are two possible values for

z

because there are two multiples of 9 that can be formed by the expression

(9+z)

where

z

is a single digit. Both

z=0

and

z=9

result in a sum of digits (9 and 18, respectively) that is divisible by 9, thus satisfying the condition.

Final Answer: The possible values for

z

are 0 and 9.

Q2Figure it Out 4

"I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.

Solution

Given: Snehal's claim. Let the first number be

N_1

. It leaves a remainder of 8 when divided by 12. So,

N_1 = 12k + 8

for some integer

k

. Let the second number be

N_2

. It is 4 short of a multiple of 12. So,

N_2 = 12m - 4

for some integer

m

.

Claim: The sum

S = N_1 + N_2

will always be a multiple of 8.

Examination: Let's find the algebraic form of the sum

S

.

S = (12k + 8) + (12m - 4)

S = 12k + 12m + 8 - 4

S = 12(k+m) + 4

Now, we need to check if

12(k+m) + 4

is always a multiple of 8. Let's test some integer values for

k

and

m

.

Case 1: Let $k=0$ and $m=1$ . $N_1 = 12(0)+8 = 8$ . $N_2 = 12(1)-4 = 8$ . Sum $S = 8+8=16$ . 16 is a multiple of 8. The claim holds here.
Case 2: Let $k=1$ and $m=1$ . $N_1 = 12(1)+8 = 20$ . $N_2 = 12(1)-4 = 8$ . Sum $S = 20+8=28$ . 28 is not a multiple of 8. The claim fails here.

Justification: The expression for the sum is

S = 12(k+m) + 4

. We can rewrite this as

S = 4(3(k+m) + 1)

. For S to be a multiple of 8, the term

(3(k+m) + 1)

must be a multiple of 2 (i.e., an even number). Let

j = k+m

. The term is

3j+1

. If

j

is odd,

3j

is odd, and

3j+1

is even. In this case, S is a multiple of 8. If

j

is even,

3j

is even, and

3j+1

is odd. In this case, S is not a multiple of 8. Since

j=k+m

can be either even or odd depending on the choice of

k

and

m

, the sum is not always a multiple of 8.

Final Answer: Snehal's claim is false. It is only sometimes true.

Q3Figure it Out 4

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Solution

Given: Two numbers, both multiples of 3.

To Find: The condition under which their sum is a multiple of 6.

Solution: Let the two multiples of 3 be

3a

and

3b

, where

a

and

b

are integers. Their sum is

S = 3a + 3b = 3(a+b)

. For the sum

S

to be a multiple of 6, it must be divisible by both 2 and 3. It is already divisible by 3. So, the condition is that

S

must be an even number. Since

S = 3(a+b)

, and 3 is odd,

S

is even if and only if

(a+b)

is even.

The sum of two integers

(a+b)

is even if and only if

a

and

b

have the same parity (both are even or both are odd).

Let's analyze the multiples of 3 themselves:

If $a$ is even, let $a=2k$ . The multiple of 3 is $3(2k) = 6k$ , which is an even multiple of 3 (i.e., a multiple of 6).
If $a$ is odd, let $a=2k+1$ . The multiple of 3 is $3(2k+1) = 6k+3$ , which is an odd multiple of 3.

So, the condition that

a

and

b

have the same parity is equivalent to the condition that the two multiples of 3 (

3a

and

3b

) have the same parity.

The different possible cases are:

Case 1: Both multiples of 3 are even. Example: $6 + 12 = 18$ . Here, 18 is a multiple of 6.
Case 2: Both multiples of 3 are odd. Example: $3 + 9 = 12$ . Here, 12 is a multiple of 6.
Case 3: One multiple of 3 is even and the other is odd. Example: $3 + 6 = 9$ . Here, 9 is not a multiple of 6.

Generalization: The sum of two multiples of 3 is a multiple of 6 if and only if both numbers are even or both numbers are odd. In other words, the sum is a multiple of 6 if the two multiples of 3 have the same parity.

Q4Figure it Out 4

Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".

(i)

Examine if her conjecture is true for any multiple of 9.

(ii)

Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Solution

(i) Examine if her conjecture is true for any multiple of 9.

Conjecture: If a number is divisible by 9, its reverse is also divisible by 9.

Examination: The divisibility rule for 9 states that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. Let's consider a number

N

. When we reverse the digits of

N

to get a new number

N_{rev}

, the set of digits in the number remains the same. Therefore, the sum of the digits of

N_{rev}

is identical to the sum of the digits of

N

.

If

N

is divisible by 9, then the sum of its digits is a multiple of 9. Since the sum of digits of

N_{rev}

is the same, it is also a multiple of 9. Therefore,

N_{rev}

is also divisible by 9.

Conclusion: Sreelatha's conjecture is Always True. Example: 198 is divisible by 9 (sum of digits = 18). Its reverse is 891. The sum of digits of 891 is

8+9+1=18

, which is divisible by 9. So 891 is divisible by 9.

(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Explanation: Yes. Similar to reversing, any shuffling (or permutation) of the digits of a number does not change the sum of the digits. If the original number is a multiple of 9, its sum of digits is a multiple of 9. Any number formed by shuffling these digits will have the same sum of digits and will, therefore, also be a multiple of 9.

Example: The number 198 is a multiple of 9. Its digits are 1, 9, 8. Other numbers formed by shuffling these digits are: 189 (sum=18, multiple of 9) 819 (sum=18, multiple of 9) 891 (sum=18, multiple of 9) 918 (sum=18, multiple of 9) 981 (sum=18, multiple of 9) All of these are multiples of 9.

Final Answer: Yes, any number formed by shuffling the digits of a multiple of 9 will also be a multiple of 9.

Q5Figure it Out 4

If $48a23b$ is a multiple of 18, list all possible pairs of values for $a$ and $b$ .

Solution

Given: The number

48a23b

is a multiple of 18.

To Find: All possible pairs of values for digits

a

and

b

.

Solution: For a number to be a multiple of 18, it must be divisible by its co-prime factors, 2 and 9.

Condition 1: Divisibility by 2 For the number to be divisible by 2, its last digit,

b

, must be an even digit. Possible values for

b

: {0, 2, 4, 6, 8}.

Condition 2: Divisibility by 9 For the number to be divisible by 9, the sum of its digits must be a multiple of 9. Sum of digits =

4 + 8 + a + 2 + 3 + b = 17 + a + b

. So,

(17 + a + b)

must be a multiple of 9.

Now we test each possible value of

b

:

If $b=0$ : The sum is $17+a+0 = 17+a$ . For this to be a multiple of 9, $17+a$ could be 18, 27, ... If $17+a=18$ , then $a=1$ . This is a valid digit. The pair is (a, b) = (1, 0).
If $b=2$ : The sum is $17+a+2 = 19+a$ . For this to be a multiple of 9, $19+a$ could be 27, 36, ... If $19+a=27$ , then $a=8$ . This is a valid digit. The pair is (a, b) = (8, 2).
If $b=4$ : The sum is $17+a+4 = 21+a$ . For this to be a multiple of 9, $21+a$ could be 27, 36, ... If $21+a=27$ , then $a=6$ . This is a valid digit. The pair is (a, b) = (6, 4).
If $b=6$ : The sum is $17+a+6 = 23+a$ . For this to be a multiple of 9, $23+a$ could be 27, 36, ... If $23+a=27$ , then $a=4$ . This is a valid digit. The pair is (a, b) = (4, 6).
If $b=8$ : The sum is $17+a+8 = 25+a$ . For this to be a multiple of 9, $25+a$ could be 27, 36, ... If $25+a=27$ , then $a=2$ . This is a valid digit. The pair is (a, b) = (2, 8).

Final Answer: The possible pairs of values for (a, b) are (1, 0), (8, 2), (6, 4), (4, 6), and (2, 8).

Q6Figure it Out 4

If $3p7q8$ is divisible by 44, list all possible pairs of values for $p$ and $q$ .

Solution

Given: The number

3p7q8

is divisible by 44.

To Find: All possible pairs of values for digits

p

and

q

.

Solution: For a number to be divisible by 44, it must be divisible by its co-prime factors, 4 and 11.

Condition 1: Divisibility by 4 For the number to be divisible by 4, the number formed by its last two digits,

q8

, must be divisible by 4. Possible values for

q8

are 08, 18(no), 28, 38(no), 48, 58(no), 68, 78(no), 88, 98(no). So, possible values for

q

are {0, 2, 4, 6, 8}.

Condition 2: Divisibility by 11 For the number to be divisible by 11, the difference between the sum of digits at odd places and the sum of digits at even places (from right) must be 0 or a multiple of 11. Number: 3 p 7 q 8 Sum of digits at odd places:

8 + 7 + 3 = 18

. Sum of digits at even places:

q + p

. Difference =

(q+p) - 18

or

18 - (p+q)

. Let's use

18 - (p+q)

. So,

18 - (p+q)

must be a multiple of 11. Possible values for

18-(p+q)

are ..., -11, 0, 11, 22, ... This means

p+q

could be

18-0=18

, or

18-11=7

, or

18-22 = -4

(not possible), or

18-(-11)=29

(not possible). So,

p+q = 7

or

p+q = 18

.

Now we combine the conditions for

q

and

p+q

:

Case 1: $p+q=7$
- If $q=0$ , then $p=7$ . The pair is (p, q) = (7, 0).
- If $q=2$ , then $p=5$ . The pair is (p, q) = (5, 2).
- If $q=4$ , then $p=3$ . The pair is (p, q) = (3, 4).
- If $q=6$ , then $p=1$ . The pair is (p, q) = (1, 6).
- If $q=8$ , then $p=-1$ . Not a valid digit.
Case 2: $p+q=18$
- If $q=8$ , then $p=10$ . Not a valid digit.
- Since $q \leq 8$ and $p \leq 9$ , the only possibility is $p=9, q=9$ . But $q$ must be in {0, 2, 4, 6, 8}. So this case yields no solutions.

Final Answer: The possible pairs of values for (p, q) are (7, 0), (5, 2), (3, 4), and (1, 6).

Q7Figure it Out 4

Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

Solution

Given: Three consecutive numbers, let them be

n, n+1, n+2

. Conditions:

$n$ is a multiple of 2.
$n+1$ is a multiple of 3.
$n+2$ is a multiple of 4.

To Find: A set of such numbers and the frequency of their occurrence.

Solution: From condition 1,

n

is even. From condition 3,

n+2

is a multiple of 4. If

n

is even, let

n=2k

. Then

n+2 = 2k+2 = 2(k+1)

. For this to be a multiple of 4,

(k+1)

must be even, which means

k

must be odd. So,

n

must be of the form

2 \times (\text{an odd number})

. Possible values for

n

: 2, 6, 10, 14, 18, 22, 26, ...

Now let's use condition 2:

n+1

is a multiple of 3. Let's test our possible values for

n

:

If $n=2$ , then $n+1=3$ . 3 is a multiple of 3. This works. The numbers are 2, 3, 4. Let's check all conditions: 2 is a multiple of 2, 3 is a multiple of 3, 4 is a multiple of 4. This is a valid set.
If $n=6$ , then $n+1=7$ . Not a multiple of 3.
If $n=10$ , then $n+1=11$ . Not a multiple of 3.
If $n=14$ , then $n+1=15$ . 15 is a multiple of 3. This works. The numbers are 14, 15, 16. Let's check: 14 is multiple of 2, 15 is multiple of 3, 16 is multiple of 4. This is another valid set.
If $n=18$ , then $n+1=19$ . Not a multiple of 3.
If $n=22$ , then $n+1=23$ . Not a multiple of 3.
If $n=26$ , then $n+1=27$ . 27 is a multiple of 3. This works. The numbers are 26, 27, 28.

Frequency of Occurrence: The first numbers of the sequences are 2, 14, 26, ... This is an arithmetic progression with a common difference of

14-2=12

and

26-14=12

. So, such sets of three consecutive numbers occur every 12 numbers.

Final Answer: One set of such numbers is 2, 3, and 4. Yes, there are more such numbers (e.g., 14, 15, 16; 26, 27, 28; etc.). They occur regularly, with the first number in the sequence being 12 more than the first number of the previous sequence.

Q8Figure it Out 4

Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Solution

To Find: Five multiples of 36 between 45,000 and 47,000.

Approach:

First, find the smallest multiple of 36 that is greater than 45,000.
To do this, divide 45,000 by 36.
Based on the remainder, find the next multiple of 36.
Once the first multiple is found, repeatedly add 36 to get the subsequent multiples.

Step-by-step Solution: Step 1 & 2: Divide 45,000 by 36.

45000 \div 36 = 1250

. The division is exact, which means 45,000 is a multiple of 36.

Step 3: Find the first multiple between 45,000 and 47,000. Since 45,000 is a multiple, the next one will be

45000 + 36 = 45036

. This is our first number.

Step 4: Find the next four multiples by adding 36 repeatedly.

First multiple: 45,036
Second multiple: $45036 + 36 =$ 45,072
Third multiple: $45072 + 36 =$ 45,108
Fourth multiple: $45108 + 36 =$ 45,144
Fifth multiple: $45144 + 36 =$ 45,180 All these numbers are between 45,000 and 47,000.

Final Answer: Five multiples of 36 between 45,000 and 47,000 are 45,036, 45,072, 45,108, 45,144, and 45,180.

Q9Figure it Out 4

The middle number in the sequence of 5 consecutive even numbers is $5p$ . Express the other four numbers in sequence in terms of $p$ .

Solution

Given: A sequence of 5 consecutive even numbers. The middle number is

5p

.

To Find: The other four numbers in terms of

p

.

Solution: Consecutive even numbers have a difference of 2 between them. The sequence has 5 numbers, so the middle number is the third number. Sequence: [1st, 2nd, 3rd, 4th, 5th]

Given that the 3rd number is

5p

.

The number before it (2nd number) will be

5p - 2

. The number before that (1st number) will be

(5p-2) - 2 = 5p - 4

.

The number after the middle one (4th number) will be

5p + 2

. The number after that (5th number) will be

(5p+2) + 2 = 5p + 4

.

So the complete sequence is:

5p-4, 5p-2, 5p, 5p+2, 5p+4

.

Final Answer: The other four numbers in the sequence are

5p-4

,

5p-2

,

5p+2

, and

5p+4

.

Q10Figure it Out 4

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Solution

Given: We need to find a 6-digit number

N

with two properties:

$N$ is divisible by 15.
The reverse of $N$ , let's call it $N_{rev}$ , is divisible by 6.

Solution: Let the 6-digit number be

N = abcdef

. Let its reverse be

N_{rev} = fedcba

.

Condition 1: $N$ is divisible by 15. This means

N

must be divisible by both 3 and 5.

Divisibility by 5: The last digit, $f$ , must be 0 or 5.
Divisibility by 3: The sum of digits, $a+b+c+d+e+f$ , must be a multiple of 3.

Condition 2: $N_{rev}$ is divisible by 6. This means

N_{rev}

must be divisible by both 2 and 3.

Divisibility by 2: The last digit of $N_{rev}$ , which is $a$ , must be an even digit. Since $a$ is the first digit of a 6-digit number, $a \neq 0$ . So, $a \in \{2, 4, 6, 8\}$ .
Divisibility by 3: The sum of digits, $f+e+d+c+b+a$ , must be a multiple of 3. This condition is the same as for $N$ .

Let's construct a number:

Choose a value for $a$ from {2, 4, 6, 8}. Let's pick a = 2.
Choose a value for $f$ from {0, 5}. Let's pick f = 5.
Now we need to choose $b, c, d, e$ such that the sum of all digits is a multiple of 3. Current sum = $a+f = 2+5 = 7$ . Sum = $7 + b+c+d+e$ . This must be a multiple of 3. Let's make a simple choice for $b, c, d$ . Let b=0, c=0, d=0. Now the sum is $7+0+0+0+e = 7+e$ . We need $7+e$ to be a multiple of 3. The smallest multiple of 3 greater than 7 is 9. If $7+e=9$ , then e=2.

Let's form the number and check:

a=2, b=0, c=0, d=0, e=2, f=5

. The number

N

is 200025. Its reverse

N_{rev}

is 520002.

Verification:

Is $N=200025$ $N = 200025$ divisible by 15?
- Ends in 5, so it's divisible by 5. (Yes)
- Sum of digits = $2+0+0+0+2+5=9$ . 9 is divisible by 3. (Yes)
- Since it's divisible by both 3 and 5, it is divisible by 15. (Correct)
Is $N_{rev}=520002$ $N_{re v} = 520002$ divisible by 6?
- Ends in 2 (even), so it's divisible by 2. (Yes)
- Sum of digits = $5+2+0+0+0+2=9$ . 9 is divisible by 3. (Yes)
- Since it's divisible by both 2 and 3, it is divisible by 6. (Correct)

Final Answer: One such 6-digit number is 200,025.

Q11Figure it Out 4

Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled". Examine if his conjecture is true; explain your conclusion.

Solution

Given: Deepak's claim about doubling multiples of 11.

Claim: Some multiples of 11, when doubled, are still multiples of 11, while others are not.

Examination: Let

N

be any multiple of 11. By definition, this means

N

can be written as

N = 11k

for some integer

k

.

Now, let's double this number: Double of

N = 2 \times N = 2 \times (11k)

. Using the associative property of multiplication, we can write this as:

2 \times (11k) = 11 \times (2k)

.

Let

m = 2k

. Since

k

is an integer,

m

is also an integer. So, the doubled number can be written as

11m

, where

m

is an integer. By definition, any number that can be written as

11 \times (\text{an integer})

is a multiple of 11.

This shows that if a number is a multiple of 11, its double will always be a multiple of 11.

Conclusion: Deepak's claim is false. There are no multiples of 11 that stop being multiples of 11 when doubled. The property holds for all multiples of 11.

Example:

22 is a multiple of 11. Doubled is 44, which is also a multiple of 11.
121 is a multiple of 11. Doubled is 242. To check 242: $(2+2)-4=0$ , so 242 is a multiple of 11.

Final Answer: Deepak's conjecture is false. If a number is a multiple of 11, its double is always a multiple of 11.

Q12Figure it Out 4

Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.

(i)

The product of a multiple of 6 and a multiple of 3 is a multiple of 9.

(ii)

The sum of three consecutive even numbers will be divisible by 6.

(iii)

If abcdef is a multiple of 6, then badcef will be a multiple of 6.

(iv)

$8(7b-3)-4(11b+1)$ is a multiple of 12.

Solution

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. Answer: Always True. Reasoning: Let the multiple of 6 be

6k

and the multiple of 3 be

3m

for some integers

k, m

. Their product is

P = (6k) \times (3m) = 18km

. We can write

18km

as

9 \times (2km)

. Since

2km

is an integer, the product

P

is always a multiple of 9.

(ii) The sum of three consecutive even numbers will be divisible by 6. Answer: Always True. Reasoning: Let the three consecutive even numbers be

2n

,

2n+2

, and

2n+4

for some integer

n

. Their sum is

S = 2n + (2n+2) + (2n+4) = 6n + 6

. We can factor out 6:

S = 6(n+1)

. Since

n+1

is an integer, the sum

S

is always a multiple of 6.

(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. Answer: Always True. Reasoning: Let

N = abcdef

. If

N

is a multiple of 6, it must be divisible by 2 and 3.

Divisibility by 2 implies the last digit, $f$ , is even.
Divisibility by 3 implies the sum of digits, $a+b+c+d+e+f$ , is a multiple of 3. Now consider the number $N' = badcef$ .
Is $N'$ divisible by 2? The last digit of $N'$ is $f$ . Since we know $f$ is even, $N'$ is divisible by 2.
Is $N'$ divisible by 3? The sum of digits of $N'$ is $b+a+d+c+e+f$ , which is the same sum as for $N$ . Since the sum is a multiple of 3, $N'$ is divisible by 3. Since $N'$ is divisible by both 2 and 3, it is always divisible by 6.

(iv) $8(7b-3)-4(11b+1)$ is a multiple of 12. Answer: Never True. Reasoning: Let's simplify the expression:

E = 8(7b-3) - 4(11b+1)

E = 56b - 24 - 44b - 4

E = (56b - 44b) - (24 + 4)

E = 12b - 28

For this expression to be a multiple of 12, both terms must be divisible by 12.

$12b$ is always a multiple of 12 for any integer $b$ .
However, 28 is not a multiple of 12 ( $28 = 12 \times 2 + 4$ ). The difference of a multiple of 12 and a non-multiple of 12 can never be a multiple of 12. Therefore, the expression is never a multiple of 12.

Q13Figure it Out 4

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Solution

To Find: The condition under which the sum of three numbers is divisible by 3.

Exploration: Any integer, when divided by 3, can have a remainder of 0, 1, or 2. Let the three numbers be

x, y, z

. Let their remainders when divided by 3 be

r_x, r_y, r_z

respectively. The sum

x+y+z

is divisible by 3 if and only if the sum of their remainders

r_x+r_y+r_z

is divisible by 3.

Possible Cases for the remainders (0, 1, 2):

Case 1: All three remainders are the same.
- (0, 0, 0): Sum of remainders = 0. Divisible by 3. (e.g., 3+6+9=18)
- (1, 1, 1): Sum of remainders = 3. Divisible by 3. (e.g., 1+4+7=12)
- (2, 2, 2): Sum of remainders = 6. Divisible by 3. (e.g., 2+5+8=15)
Case 2: All three remainders are different.
- (0, 1, 2): Sum of remainders = 3. Divisible by 3. (e.g., 3+4+5=12)
Case 3: Two remainders are the same, one is different.
- (0, 0, 1): Sum = 1. Not divisible by 3. (e.g., 3+6+4=13)
- (0, 0, 2): Sum = 2. Not divisible by 3. (e.g., 3+6+5=14)
- (1, 1, 0): Sum = 2. Not divisible by 3. (e.g., 1+4+3=8)
- (1, 1, 2): Sum = 4. Not divisible by 3. (e.g., 1+4+5=10)
- (2, 2, 0): Sum = 4. Not divisible by 3. (e.g., 2+5+3=10)
- (2, 2, 1): Sum = 5. Not divisible by 3. (e.g., 2+5+4=11)

Generalization: The sum of three integers is divisible by 3 if and only if one of the following conditions is met regarding their remainders upon division by 3:

All three numbers have the same remainder.
The three numbers have distinct remainders (one has remainder 0, one has remainder 1, and one has remainder 2).

Q14Figure it Out 4

Is the product of two consecutive integers always multiple of 2? Why? What about the product of three consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Solution

Product of two consecutive integers: Let the integers be

n

and

n+1

. In any pair of consecutive integers, one must be even. An even number is a multiple of 2. Therefore, their product will always have a factor of 2. Answer: Yes, it is always a multiple of 2.

Product of three consecutive integers: Let the integers be

n, n+1, n+2

. For their product to be a multiple of 6, it must be a multiple of 2 and 3.

Divisibility by 2: In any three consecutive integers, at least one is even. So the product is a multiple of 2.
Divisibility by 3: In any set of three consecutive integers, exactly one must be a multiple of 3. Since the product is a multiple of both 2 and 3 (which are co-prime), it must be a multiple of $2 \times 3 = 6$ . Answer: Yes, it is always a multiple of 6.

Product of four consecutive integers: Let the integers be

n, n+1, n+2, n+3

.

The product contains at least one multiple of 3.
The product contains two even numbers. One of these two even numbers must be a multiple of 4. So the product has factors of 3 and 4 and another even number, which means it is divisible by $3 \times 4 \times 2 = 24$ . Answer: The product of four consecutive integers is always a multiple of 24.

Product of five consecutive integers: Let the integers be

n, n+1, n+2, n+3, n+4

.

The product contains exactly one multiple of 5.
It contains at least one multiple of 4.
It contains at least one multiple of 3.
It contains at least two even numbers. The product is divisible by 5, 4, and 3. Since these are co-prime, it is divisible by their product $5 \times 4 \times 3 = 60$ . In fact, it is always divisible by $5! = 120$ . Answer: The product of five consecutive integers is always a multiple of 120.

Q15Figure it Out 4

Solve the cryptarithms -

(i)

EF $\times$ E = GGG

(ii)

WOW $\times$ 5 = MEOW

Solution

(i) EF $\times$ E = GGG

Given:

(10E + F) \times E = 111G = 3 \times 37 \times G

.

Solution: The right side has a prime factor of 37. Therefore, one of the factors on the left side, either E or EF, must be a multiple of 37.

E cannot be a multiple of 37 as it is a single digit.
Therefore, EF (a two-digit number) must be a multiple of 37. Possible multiples of 37 are $37 \times 1 = 37$ and $37 \times 2 = 74$ .

Case 1: EF = 37 This means E=3 and F=7. The equation becomes

37 \times 3 = GGG

.

37 \times 3 = 111

. So, GGG = 111, which means G=1. This solution works: E=3, F=7, G=1. All letters are different digits.

Case 2: EF = 74 This means E=7 and F=4. The equation becomes

74 \times 7 = GGG

.

74 \times 7 = 518

. This is not of the form GGG.

So, the only solution is from Case 1.

Final Answer for (i): E=3, F=7, G=1.

(ii) WOW $\times$ 5 = MEOW

Given:

(100W + 10O + W) \times 5 = 1000M + 100E + 10O + W

.

Solution:

From the units place, the product $W \times 5$ must end in the digit W. This only happens if W=0 or W=5.
- W cannot be 0, because it is the first digit of WOW.
- Therefore, W=5.
The puzzle is now: $5O5 \times 5 = ME O 5$ . The product MEOW is a 4-digit number. $5O5 \times 5$ is roughly $500 \times 5 = 2500$ . So M must be 2. M=2.
The puzzle is now: $5O5 \times 5 = 2E O 5$ . Let's perform the multiplication: $(500 + 10O + 5) \times 5 = 2500 + 50O + 25 = 2525 + 50O$ . The result is $2EO5 = 2000 + 100E + 10O + 5$ . So, $2525 + 50O = 2005 + 100E + 10O$ . $520 + 40O = 100E$ . Divide by 20: $26 + 2O = 5E$ .
We need to find digits O and E that satisfy $5E = 26 + 2O$ . Also, O and E must be different from W=5 and M=2. The right side, $26+2O$ , must be a multiple of 5. This means it must end in 0 or 5.
- $26+2O$ is always even. So it must end in 0.
- For $26+2O$ to end in 0, $2O$ must end in 4. This happens if O=2 or O=7.

Case a: O=2 If O=2, then

5E = 26 + 2(2) = 30

. So, E=6. Let's check the digits: W=5, O=2, M=2, E=6. Here M and O are the same digit (2). This is allowed in some cryptarithms. Let's check the multiplication:

525 \times 5 = 2625

. This fits the pattern MEOW = 2625. This is a valid solution.

Case b: O=7 If O=7, then

5E = 26 + 2(7) = 26 + 14 = 40

. So, E=8. Let's check the digits: W=5, O=7, M=2, E=8. All are different. Let's check the multiplication:

575 \times 5 = 2875

. This fits the pattern MEOW = 2875. This is also a valid solution.

Typically, if not specified, different letters mean different digits. If so, Case b is the better answer. If letters can represent the same digit, both are correct. Let's assume different letters must be different digits. Checking again Case a: M=2, O=2. This is allowed if the problem allows it. Let's check the original puzzle again WOW. This implies the two Ws are the same. MEOW implies M,E,O,W can be different. The letter O appears in both WOW and MEOW, so it must be the same digit in both places. In solution a, O is 2 everywhere. It works. Let's provide the solution where all letters are distinct if possible, which is Case b. Wait, in case (a), M=2 and O=2. These are different letters. So they must be different digits. So case (a) is invalid. Case (b) gives W=5, O=7, M=2, E=8. All different. This is the correct solution.

Let's re-verify my calculation for case (a). O=2.

5E=30, E=6

. W=5, M=2, O=2, E=6. M and O are different letters but have the same value 2. So this is not a valid solution. My previous statement was wrong. Therefore only case (b) is valid.

Final Answer for (ii): W=5, O=7, M=2, E=8. (i.e.,

575 \times 5 = 2875

)

Q16Figure it Out 4

Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Solution

To Find: The Venn diagram representing the relationship between multiples of 4, 8, and 32.

Analysis of the relationship:

A multiple of 32 can be written as $32k = 8 \times (4k)$ . This means every multiple of 32 is also a multiple of 8.
A multiple of 8 can be written as $8m = 4 \times (2m)$ . This means every multiple of 8 is also a multiple of 4.

From this, we can conclude:

The set of multiples of 32 is a subset of the set of multiples of 8.
The set of multiples of 8 is a subset of the set of multiples of 4.

This relationship is represented by three concentric circles, where the innermost circle represents the smallest set (multiples of 32), the middle circle represents the next set (multiples of 8), and the outermost circle represents the largest set (multiples of 4).

Conclusion: The correct Venn diagram is the one with three nested circles, one inside the other.

Final Answer: The correct diagram is (iii).

Q1Fill in the following table

Fill in the following table. Find a quick way to do this?
Number 2 3 4 5 6 8 9 10 11
128 Yes No No No No Yes No No No
990
1586
275
6686
639210
429714
2856
3060
406839

Solution

Quick Way (Divisibility Rules):

Divisible by 2: Last digit is even (0, 2, 4, 6, 8).
Divisible by 3: Sum of digits is divisible by 3.
Divisible by 4: Number formed by last two digits is divisible by 4.
Divisible by 5: Last digit is 0 or 5.
Divisible by 6: Divisible by both 2 and 3.
Divisible by 8: Number formed by last three digits is divisible by 8.
Divisible by 9: Sum of digits is divisible by 9.
Divisible by 10: Last digit is 0.
Divisible by 11: Difference between the sum of digits at odd places and the sum of digits at even places (from right) is 0 or a multiple of 11.

Completed Table:

Number	2	3	4	5	6	8	9	10	11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	Yes	Yes	No	Yes	Yes	No	Yes	Yes	Yes
1586	Yes	No	No	No	No	No	No	No	No
275	No	No	No	Yes	No	No	No	No	Yes
6686	Yes	No	No	No	No	No	No	No	No
639210	Yes	Yes	No	Yes	Yes	No	Yes	Yes	Yes
429714	Yes	Yes	No	No	Yes	No	No	No	No
2856	Yes	Yes	Yes	No	Yes	Yes	No	No	No
3060	Yes	Yes	Yes	Yes	Yes	No	Yes	Yes	No
406839	No	Yes	No	No	No	No	No	No	No

Q1Solve the following (Cryptarithms)

(i) UT $\times$ 3 = PUT

Solution

Given: UT

\times

3 = PUT This represents

(10 \times U + T) \times 3 = 100 \times P + 10 \times U + T

.

Solution: Since PUT is a 3-digit number, P cannot be 0. Let's analyze the multiplication in columns. U T $\times$ 3

P U T

From the units column,

3 \times T

must end in the digit T. Let's see which digits work for T:

3 \times 0 = 0

. (T=0)

3 \times 1 = 3

.

3 \times 2 = 6

.

3 \times 3 = 9

.

3 \times 4 = 12

.

3 \times 5 = 15

. (T=5)

3 \times 6 = 18

.

3 \times 7 = 21

.

3 \times 8 = 24

.

3 \times 9 = 27

. So, T must be 0 or 5.

Case 1: T = 0 The equation becomes U0

\times

3 = PU0.

(10 \times U) \times 3 = 100 \times P + 10 \times U

.

30U = 100P + 10U

.

20U = 100P

.

U = 5P

. Since U is a single digit, P can only be 1. If P=1, then U=5. Let's check:

50 \times 3 = 150

. So, U=5, T=0, P=1. This works.

Case 2: T = 5 The equation becomes U5

\times

3 = PU5. From the units column,

3 \times 5 = 15

. This means we write down 5 and carry over 1. From the tens column,

(3 \times U) + 1

(carry-over) must end in the digit U. Let's test values for U:

U=1: (3 \times 1) + 1 = 4

.

U=2: (3 \times 2) + 1 = 7

. ...

U=8: (3 \times 8) + 1 = 25

. This ends in 5, not U=8. Let's write it as an equation:

3U+1 = 10k + U

for some integer k.

2U+1 = 10k

.

2U+1

must be a multiple of 10. The only single-digit U for which this is true is not possible. For example if U=1,

2(1)+1=3

. If U=9,

2(9)+1=19

. No single digit U works.

So only Case 1 is valid.

Final Answer: P=1, U=5, T=0.

Q2Solve the following (Cryptarithms)

(ii) AB $\times$ 5 = BC

Solution

Given: AB

\times

5 = BC This represents

(10A + B) \times 5 = 10B + C

.

Solution: The product BC is a 2-digit number, so

AB < 20

. Thus, A must be 1. (A cannot be 0). So the puzzle is 1B

\times

5 = BC.

(10 + B) \times 5 = 10B + C

.

50 + 5B = 10B + C

.

50 = 5B + C

.

Also, from the multiplication, the product of

5 \times B

must end in C. And the product of

5 \times 1

plus any carry-over must equal B. Let's test values for B: If B=5,

5 \times 5 = 25

. So C=5, carry-over is 2. Then in the tens place,

(5 \times 1) + 2 = 7

. This should be B. But

B=5

, so

7 \neq 5

. Does not work. If B=6,

5 \times 6 = 30

. So C=0, carry-over is 3. Then in the tens place,

(5 \times 1) + 3 = 8

. This should be B. But

B=6

, so

8 \neq 6

. Does not work. Let's use the equation

50 = 5B + C

. Since C is a digit (

0 \leq C \leq 9

),

5B

must be between 41 and 50. This means B must be 9. (

5 \times 9 = 45

). If B=9, then

50 = 5(9) + C \implies 50 = 45 + C \implies C=5

. Let's check this solution: A=1, B=9, C=5. AB = 19. BC = 95.

19 \times 5 = 95

. This is correct.

Final Answer: A=1, B=9, C=5.

Q3Solve the following (Cryptarithms)

(iii) L2N $\times$ 2 = 2NP

Solution

Given: L2N

\times

2 = 2NP This represents

(100L + 20 + N) \times 2 = 200 + 10N + P

.

Solution: Let's write it in column format: L 2 N $\times$ 2

2 N P

From the hundreds column,

2 \times L

(plus any carry-over from the tens column) must equal 2. Since L is the first digit,

L \neq 0

. This implies

L

must be 1, and there is no carry-over from the tens place. So, L=1.

For there to be no carry-over from the tens place,

2 \times 2

(plus any carry-over from units) must be less than 10. The value is 4. So

4 + \text{carry} < 10

. This is true. From the tens column,

2 \times 2 + \text{carry from units} = N

.

From the units column,

2 \times N

results in a number ending in P. Let's call the carry-over to the tens place

c_1

. So,

2N = 10c_1 + P

.

From the tens column,

2 \times 2 + c_1 = N

.

4 + c_1 = N

.

Substitute N in the first equation:

2(4+c_1) = 10c_1 + P

8 + 2c_1 = 10c_1 + P

8 = 8c_1 + P

.

Since P is a digit (

P \geq 0

), and

c_1

is a carry-over from

2 \times N

,

c_1

can only be 0 or 1. (Maximum

2 \times 9 = 18

, so max carry is 1). If

c_1=1

, then

8 = 8(1) + P \implies 8 = 8 + P \implies P=0

. If

c_1=0

, then

8 = 8(0) + P \implies P=8

.

Let's check both cases for

c_1

: Case 1: $c_1 = 1$ and P = 0. If

c_1=1

, then

N = 4 + c_1 = 4+1=5

. Let's check if

2 \times N

gives a carry-over of 1.

2 \times 5 = 10

. The last digit is 0 (which is P), and the carry-over is 1. This works. So, L=1, N=5, P=0. Let's check the multiplication:

125 \times 2 = 250

. This matches 2NP with N=5, P=0.

Case 2: $c_1 = 0$ and P = 8. If

c_1=0

, then

N = 4 + c_1 = 4+0=4

. Let's check if

2 \times N

gives a carry-over of 0.

2 \times 4 = 8

. The last digit is 8 (which is P), and the carry-over is 0. This works. So, L=1, N=4, P=8. Let's check the multiplication:

124 \times 2 = 248

. This matches 2NP with N=4, P=8. This solution is also valid. However, in cryptarithms, different letters usually represent different digits. In this solution, L=1, N=4, P=8. All are different. In the first solution, L=1, N=5, P=0. All are different. The problem does not state that there is a unique solution. Let's re-read the source. The source does not specify a unique solution. I will provide the first one found. Let's re-check the problem statement: L2N x 2 = 2NP. It seems the problem is written this way. Both solutions are mathematically sound. Let's stick with the first one.

Final Answer: L=1, N=5, P=0.

Q4Solve the following (Cryptarithms)

(iv) XY $\times$ 4 = ZX

Solution

Given: XY

\times

4 = ZX This represents

(10X + Y) \times 4 = 10Z + X

.

Solution: Since ZX is a 2-digit number,

XY \times 4 < 100

, so

XY < 25

. This means X can be 1 or 2. X cannot be 0.

Case 1: X = 1 The puzzle is 1Y

\times

4 = Z1. From the units column,

4 \times Y

must end in 1. Let's check the multiplication table of 4:

4 \times 1 = 4

4 \times 2 = 8

4 \times 3 = 12

4 \times 4 = 16

4 \times 5 = 20

4 \times 6 = 24

4 \times 7 = 28

4 \times 8 = 32

4 \times 9 = 36

No product ends in 1. So X cannot be 1.

Case 2: X = 2 The puzzle is 2Y

\times

4 = Z2. From the units column,

4 \times Y

must end in 2. From the table above, this happens when Y=3 or Y=8.

Subcase 2a: Y = 3 Let's do the multiplication:

23 \times 4 = 92

. This fits the pattern Z2, where Z=9 and X=2. Let's check the letters: X=2, Y=3, Z=9. All are different. This is a valid solution.

Subcase 2b: Y = 8 Let's do the multiplication:

28 \times 4 = 112

. This is a 3-digit number, but ZX is a 2-digit number. So this case is not possible.

The only solution is X=2, Y=3, Z=9.

Final Answer: X=2, Y=3, Z=9.

Q5Solve the following (Cryptarithms)

(v) PP $\times$ QQ = PRP

Solution

Given: PP

\times

QQ = PRP This represents

(11 \times P) \times (11 \times Q) = 101 \times P + 10 \times R

.

121 \times P \times Q = 101P + 10R

.

Solution: Let's analyze the multiplication structure. P P $\times$ Q Q

P P Q (This is

PP \times Q

)

P P Q 0 (This is $PP \times Q \times 10$ )

P R P

The sum is

(PP \times Q) + (PP \times Q0) = (PP \times Q) \times (1+10) = (PP \times Q) \times 11

. So,

PRP = (PP \times Q) \times 11

. PRP is a multiple of 11. The divisibility rule for 11 for a 3-digit number PRP is

P - R + P = 0

or a multiple of 11. So,

2P - R = 0

or

2P - R = 11

.

Case 1: $2P - R = 0 \implies R = 2P$ . Let's test values for P. P cannot be 0. If P=1, R=2. PRP = 121.

121 = (11 \times Q) \times 11 = 121 \times Q

. This implies Q=1. But P and Q must be different. Let's assume they can be the same.

11 \times 11 = 121

. PP=11, QQ=11, PRP=121. P=1, Q=1, R=2. This works if P and Q can be the same. If P=2, R=4. PRP = 242.

242 = (22 \times Q) \times 11 = 242 \times Q

. Implies Q=1. Check:

22 \times 11 = 242

. So PP=22, QQ=11, PRP=242. P=2, Q=1, R=4. This is a valid solution. If P=3, R=6. PRP = 363.

363 = (33 \times Q) \times 11 = 363 \times Q

. Implies Q=1. Check:

33 \times 11 = 363

. So PP=33, QQ=11, PRP=363. P=3, Q=1, R=6. This is a valid solution. If P=4, R=8. PRP = 484.

484 = (44 \times Q) \times 11 = 484 \times Q

. Implies Q=1. Check:

44 \times 11 = 484

. So PP=44, QQ=11, PRP=484. P=4, Q=1, R=8. This is a valid solution. If P=5, R=10. R must be a single digit. This case is not possible.

Case 2: $2P - R = 11$ . Since R is a digit (

0 \leq R \leq 9

), let's see possible values for P. If P=6,

12 - R = 11 \implies R=1

. PRP = 616.

616 = (66 \times Q) \times 11

.

56 = 66 \times Q

. Q is not an integer. If P=7,

14 - R = 11 \implies R=3

. PRP = 737.

737 = (77 \times Q) \times 11

.

67 = 77 \times Q

. Q is not an integer. If P=8,

16 - R = 11 \implies R=5

. PRP = 858.

858 = (88 \times Q) \times 11

.

78 = 88 \times Q

. Q is not an integer. If P=9,

18 - R = 11 \implies R=7

. PRP = 979.

979 = (99 \times Q) \times 11

.

89 = 99 \times Q

. Q is not an integer.

It seems the only valid solutions come from Case 1 with Q=1. Let's pick one solution, for example P=2, Q=1, R=4.

Final Answer: P=2, Q=1, R=4. (i.e.,

22 \times 11 = 242

)

Q6Solve the following (Cryptarithms)

(vi) JK $\times$ 6 = KKK

Solution

Given: JK

\times

6 = KKK This represents

(10J + K) \times 6 = 100K + 10K + K = 111K

.

Solution:

111K = 3 \times 37 \times K

. So,

(10J + K) \times 6 = 3 \times 37 \times K

. Divide both sides by 3:

(10J + K) \times 2 = 37 \times K

.

20J + 2K = 37K

.

20J = 35K

. Divide both sides by 5:

4J = 7K

.

Since J and K are single non-zero digits (K cannot be 0 because KKK is a 3-digit number, and J cannot be 0 as it's the first digit of JK), and 4 and 7 are coprime, the only possible solution is: J = 7 and K = 4.

Let's check this solution: JK = 74. KKK = 444.

74 \times 6 = 444

. This is correct.

Final Answer: J=7, K=4.

Number PlayClass 8 Mathematics NCERT Solutions

The sum of four consecutive numbers is 34. What are these numbers?

Solution

Suppose ppp is the greatest of five consecutive numbers. Describe the other four numbers in terms of ppp.

Solution

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

The sum of two even numbers is a multiple of 3.

If a number is not divisible by 18, then it is also not divisible by 9.

If two numbers are not divisible by 6, then their sum is not divisible by 6.

The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Solution

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Solution

Solution

Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tathagat's claim true?

Solution

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.

4779+6614779+6614779+661

4779-661

Solution

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Solution

Find, without dividing, whether the following numbers are divisible by 9.

123

405

8888

93547

358095

Solution

Find the smallest multiple of 9 with no odd digits.

Solution

Find the multiple of 9 that is closest to the number 6000.

Solution

How many multiples of 9 are there between the numbers 4300 and 4400?

Solution

The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

Solution

Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Solution

What will be the digital root of the number 9a+36b+139a+36b+139a+36b+13?

Solution

Make conjectures by examining if there are any patterns or relations between

the parity of a number and its digital root.

the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Solution

If 31z531z531z5 is a multiple of 9, where zzz is a digit, what is the value of zzz? Explain why there are two answers to this problem.

Solution

"I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.

Solution

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Solution

Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".

Examine if her conjecture is true for any multiple of 9.

Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Solution

If 48a23b48a23b48a23b is a multiple of 18, list all possible pairs of values for aaa and bbb.

Solution

If 3p7q83p7q83p7q8 is divisible by 44, list all possible pairs of values for ppp and qqq.

Solution

Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

Solution

Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Solution

The middle number in the sequence of 5 consecutive even numbers is 5p5p5p. Express the other four numbers in sequence in terms of ppp.

Solution

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Solution

Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled". Examine if his conjecture is true; explain your conclusion.

Solution

Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.

The product of a multiple of 6 and a multiple of 3 is a multiple of 9.

The sum of three consecutive even numbers will be divisible by 6.

If abcdef is a multiple of 6, then badcef will be a multiple of 6.

8(7b−3)−4(11b+1)8(7b-3)-4(11b+1)8(7b−3)−4(11b+1) is a multiple of 12.

Solution

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Solution

Is the product of two consecutive integers always multiple of 2? Why? What about the product of three consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Solution

Suppose $p$ is the greatest of five consecutive numbers. Describe the other four numbers in terms of $p$ .

$4779+661$

What will be the digital root of the number $9a+36b+13$ ?

If $31z5$ is a multiple of 9, where $z$ is a digit, what is the value of $z$ ? Explain why there are two answers to this problem.

If $48a23b$ is a multiple of 18, list all possible pairs of values for $a$ and $b$ .

If $3p7q8$ is divisible by 44, list all possible pairs of values for $p$ and $q$ .

The middle number in the sequence of 5 consecutive even numbers is $5p$ . Express the other four numbers in sequence in terms of $p$ .

$8(7b-3)-4(11b+1)$ is a multiple of 12.

EF $\times$ E = GGG

WOW $\times$ 5 = MEOW

Fill in the following table. Find a quick way to do this?
Number 2 3 4 5 6 8 9 10 11
128 Yes No No No No Yes No No No
990
1586
275
6686
639210
429714
2856
3060
406839

(i) UT $\times$ 3 = PUT

Solution: Since PUT is a 3-digit number, P cannot be 0. Let's analyze the multiplication in columns. U T $\times$ 3

(ii) AB $\times$ 5 = BC

(iii) L2N $\times$ 2 = 2NP

Solution: Let's write it in column format: L 2 N $\times$ 2

(iv) XY $\times$ 4 = ZX

(v) PP $\times$ QQ = PRP

Solution: Let's analyze the multiplication structure. P P $\times$ Q Q

(vi) JK $\times$ 6 = KKK