Prime TimeClass 6 Mathematics NCERT Solutions

Given: In the 'idli-vada' game, 'idli' is said for multiples of 3, 'vada' for multiples of 5, and 'idli-vada' for common multiples of 3 and 5.

To Find: The number at which 'idli-vada' is said for the 10th time.

Solution: 'Idli-vada' is said for numbers that are common multiples of both 3 and 5. The first common multiple is the Least Common Multiple (LCM) of 3 and 5. Since 3 and 5 are prime numbers, their LCM is their product. $\text{LCM}(3, 5) = 3 \times 5 = 15$ The numbers where 'idli-vada' is said are the multiples of 15: 15, 30, 45, 60, and so on. The 10th time 'idli-vada' is said will be at the 10th multiple of 15. $10 \times 15 = 150$

Final Answer: 'Idli-vada' is said for the 10th time at the number 150.

Given: The game is played for numbers from 1 to 90. 'idli' is for multiples of 3. 'vada' is for multiples of 5. 'idli-vada' is for common multiples of 3 and 5 (i.e., multiples of 15).

a. Number of times 'idli' is said: This is the number of multiples of 3 from 1 to 90. To find this, we divide 90 by 3. $\text{Number of multiples of 3} = \frac{90}{3} = 30$ So, 'idli' would be said 30 times.

b. Number of times 'vada' is said: This is the number of multiples of 5 from 1 to 90. To find this, we divide 90 by 5. $\text{Number of multiples of 5} = \frac{90}{5} = 18$ So, 'vada' would be said 18 times.

c. Number of times 'idli-vada' is said: This is the number of common multiples of 3 and 5, which are multiples of their LCM, 15. To find this, we divide 90 by 15. $\text{Number of multiples of 15} = \frac{90}{15} = 6$ So, 'idli-vada' would be said 6 times.

Final Answer: a. 'idli' would be said 30 times. b. 'vada' would be said 18 times. c. 'idli-vada' would be said 6 times.

Given: The game is played for numbers from 1 to 900.

Solution: We follow the same logic as in the previous question, but with the upper limit of 900.

a. Number of times 'idli' is said (multiples of 3): $\frac{900}{3} = 300$

b. Number of times 'vada' is said (multiples of 5): $\frac{900}{5} = 180$

c. Number of times 'idli-vada' is said (multiples of 15): $\frac{900}{15} = 60$

Final Answer: If the game was played till 900:

Explanation: Yes, the figure (a Venn diagram) is related to the 'idli-vada' game. It visually represents the multiples of the two numbers involved.

Figure for the game played till 60: Let's consider the original game with numbers 3 and 5, played up to 60.

This shows that the Venn diagram is a useful tool to visualize the concepts of multiples and common multiples as used in the game.

To Find: Multiples of 40 between 310 and 410.

Solution: We can find the multiples of 40 by multiplying 40 by integers. $40 \times 7 = 280$ (too small) $40 \times 8 = 320$. This is between 310 and 410. $40 \times 9 = 360$. This is between 310 and 410. $40 \times 10 = 400$. This is between 310 and 410. $40 \times 11 = 440$ (too large)

Final Answer: The multiples of 40 that lie between 310 and 410 are 320, 360, and 400.

Final Answer: a. 35 b. 45

Given: A perfect number is a number for which the sum of all its factors is equal to twice the number.

To Find: A perfect number between 1 and 10.

Solution: Let's check each number from 1 to 10.

Final Answer: The perfect number between 1 and 10 is 6.

Final Answer: a. Common factors of 20 and 28 are 1, 2, 4. b. Common factors of 35 and 50 are 1, 5. c. Common factors of 4, 8 and 12 are 1, 2, 4. d. Common factors of 5, 15 and 25 are 1, 5.

Solution: A number that is a multiple of 25 but not a multiple of 50 must be an odd multiple of 25. Let's list the multiples of 25:

We can choose any three from the numbers that are not multiples of 50.

Final Answer: Three such numbers are 25, 75, and 125.

To Find: Possible pairs of numbers ($a, b$).

Solution: We need to find pairs of numbers less than 10 whose LCM is greater than 50. Let's try some pairs:

Final Answer: The two numbers could be the pair (7, 8), or (8, 9), or (7, 9).

Given: Treasures are placed at numbers 28 and 70.

To Find: Jump sizes that will land on both numbers.

Solution: The jump sizes that land on both numbers must be common factors of 28 and 70.

First, find the factors of 28: Factors of 28 are {1, 2, 4, 7, 14, 28}.

Next, find the factors of 70: Factors of 70 are {1, 2, 5, 7, 10, 14, 35, 70}.

Now, find the common factors: The common factors of 28 and 70 are {1, 2, 7, 14}.

Final Answer: The jump sizes that will land on both numbers are 1, 2, 7, and 14.

Given: A Venn diagram where only the intersection is filled with the numbers 12, 24, 36.

Solution: The numbers in the intersection are the common multiples of two numbers. The numbers 12, 24, 36 are all multiples of 12. This means the Least Common Multiple (LCM) of the two original numbers is 12. We need to find two numbers whose LCM is 12. There are several possibilities, for example:

Let's choose the pair (3, 4) as a possible solution.

This fits the given information.

Final Answer: The two numbers could be 3 and 4. The first circle would contain Multiples of 3, and the second circle would contain Multiples of 4.

To Find: The smallest number that is a multiple of {1, 2, 3, 4, 5, 6, 8, 9, 10}. This is the Least Common Multiple (LCM) of these numbers.

Solution: To find the LCM, we use the prime factorization of each number. $1 = 1$ $2 = 2$ $3 = 3$ $4 = 2^2$ $5 = 5$ $6 = 2 \times 3$ $8 = 2^3$ $9 = 3^2$ $10 = 2 \times 5$

The LCM is the product of the highest power of each prime factor present in the numbers.

LCM = $2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 72 \times 5 = 360$.

Final Answer: The smallest number is 360.

To Find: The smallest number that is a multiple of all numbers from 1 to 10, i.e., LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10).

Solution: From the previous question, we know that LCM(1, 2, 3, 4, 5, 6, 8, 9, 10) = 360. Now we need to include the number 7. The required number is LCM(360, 7).

The prime factorization of 360 is $2^3 \times 3^2 \times 5$. The prime factorization of 7 is just 7. Since 360 and 7 have no common prime factors (they are co-prime), their LCM is their product.

LCM(360, 7) = $360 \times 7 = 2520$.

Final Answer: The smallest number is 2520.

Given: The game is played up to 60 with three different pairs of numbers.

Solution: We describe the contents of the Venn diagram for each pair.

Given: A Venn diagram shows multiples of a number in one circle and common multiples in the intersection. The multiples of the first number are 6, 12, 18, 24, 30, ... This means the first number is 6. The common multiples shown are 24, 48, 72, ... These are multiples of 24. This means the Least Common Multiple (LCM) of 6 and the other number is 24.

To Find: The other number from the given options.

Solution: Let the other number be $x$. We are looking for $x$ such that LCM(6, $x$) = 24. Let's check each option:

Final Answer: The other number could be 8.

Given: Treasures are placed at numbers 15 and 30.

To Find: All possible jump sizes that will land on both 15 and 30.

Solution: A jump size can reach a number if the number is a multiple of the jump size. This means the jump size must be a factor of the number. To reach both 15 and 30, the jump size must be a common factor of 15 and 30.

First, find the factors of 15: Factors of 15 are {1, 3, 5, 15}.

Next, find the factors of 30: Factors of 30 are {1, 2, 3, 5, 6, 10, 15, 30}.

Now, find the common factors: The common factors of 15 and 30 are {1, 3, 5, 15}.

Final Answer: The possible jump sizes that can reach both 15 and 30 are 1, 3, 5, and 15.

Given: A table of numbers from 31 to 70 with some numbers shaded and some circled.

1. Common among shaded numbers: The question intends for the shaded numbers to be multiples of 3. (Note: The shading in the provided image may not perfectly match this intention). Based on the structure of the problem, the shaded numbers are the multiples of 3 in the range 31 to 70. These are: 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69. So, the common property is that they are all multiples of 3.

2. Common among circled numbers: The circled numbers are 32, 36, 40, 44, 48, 52, 56, 60, 64, 68. Let's check for a common property. Dividing each by 4: $32 \div 4 = 8$ $36 \div 4 = 9$ $40 \div 4 = 10$ ...and so on. All circled numbers are multiples of 4.

3. Numbers both shaded and circled: These are numbers that are multiples of both 3 and 4. They are common multiples of 3 and 4. Looking at the lists, the numbers that appear in both are: 36, 48, 60. These numbers are called common multiples of 3 and 4.

Answer: No, there is no other even prime number.

Reason: A prime number is a number greater than 1 that has only two factors: 1 and itself. An even number is any integer that is divisible by 2. Any even number greater than 2 can be written as $2 \times n$, where $n$ is an integer greater than 1. This means that any even number greater than 2 has at least three factors: 1, 2, and the number itself. Therefore, it cannot be a prime number. The number 2 is the only even prime number.

Given: The list of prime numbers up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Solution: Smallest difference: We calculate the difference between consecutive primes: $3 - 2 = 1$ $5 - 3 = 2$ $7 - 5 = 2$ ... The smallest difference is 1, which occurs between 2 and 3.

Largest difference: Let's check the larger gaps: $29 - 23 = 6$ $89 - 83 = 6$ $97 - 89 = 8$ The largest difference is 8, which occurs between 89 and 97.

Final Answer: The smallest difference is 1 (between 2 and 3). The largest difference is 8 (between 89 and 97).

Solution: Let's count the number of primes in each decade (row in the table):

Final Answer: No, the number of primes is not equal in every decade. The decade 91-100 has the least number of primes (one). The decades 1-10 and 11-20 have the most number of primes (four each).

Final Answer: The prime numbers are 23 and 37.

Given: Prime numbers less than 20 are: 2, 3, 5, 7, 11, 13, 17, 19.

To Find: Three pairs from this list whose sum is a multiple of 5.

Solution: Let's test some pairs:

We can choose any three of these pairs.

Final Answer: Three such pairs are (2, 3), (3, 7), and (2, 13).

To Find: Pairs of two-digit prime numbers up to 100 where the digits are reversed.

Solution: We are looking for pairs $(p, q)$ where $p$ and $q$ are prime, and if $p$ has digits $ab$, then $q$ has digits $ba$.

Other pairs like (19, 91) do not work because 91 is not prime ($91 = 7 \times 13$).

Final Answer: The pairs are (17, 71), (37, 73), and (79, 97).

To Find: A sequence of seven consecutive numbers between 1 and 100 that are all composite.

Solution: We need to find a gap of at least 8 between two consecutive prime numbers. Let's look at the list of primes: ..., 83, 89, 97, ... The gap between 89 and 97 is $97 - 89 = 8$. The numbers between them are 90, 91, 92, 93, 94, 95, 96. There are seven numbers in this sequence. Let's check if they are all composite:

Final Answer: Seven consecutive composite numbers between 1 and 100 are 90, 91, 92, 93, 94, 95, 96.

Given: Twin primes are pairs of prime numbers $(p, p+2)$.

To Find: All twin prime pairs between 1 and 100.

Solution: We look for pairs of primes with a difference of 2 from the list of primes up to 100.

Final Answer: The twin prime pairs between 1 and 100 are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), and (71, 73).

a. There is no prime number whose units digit is 4. Answer: True. Reason: Any number with a units digit of 4 is an even number. The only even prime number is 2. All other even numbers are greater than 2 and divisible by 2, making them composite.

b. A product of primes can also be prime. Answer: False. Reason: The product of two or more prime numbers will have those primes as its factors (in addition to 1 and the product itself). Since it has more than two factors, it is a composite number. For example, $2 \times 3 = 6$, which is composite.

c. Prime numbers do not have any factors. Answer: False. Reason: Prime numbers have exactly two factors: 1 and the number itself.

d. All even numbers are composite numbers. Answer: False. Reason: The number 2 is an even number, but it is a prime number.

e. 2 is a prime and so is the next number, 3. For every other prime, the next number is composite. Answer: True. Reason: Every prime number other than 2 is an odd number. If $p$ is a prime number and $p > 2$, then $p$ is odd. The next number is $p+1$. Since $p$ is odd, $p+1$ must be an even number. Since $p > 2$, $p+1$ will be an even number greater than 2, which means it is composite.

To Find: The number that is a product of three different prime numbers.

Solution: We find the prime factorization of each number.

Final Answer: The number that is the product of exactly three distinct prime numbers is 105.

Given: Digits to use are 2, 4, and 5, each used once.

Solution: Let's list all possible three-digit numbers we can form: 245, 254, 425, 452, 524, 542.

Now, let's check if any of them are prime:

All the numbers that can be formed are composite.

Final Answer: None. 0 three-digit prime numbers can be made.

To Find: Five prime numbers $p$ such that $2p+1$ is also a prime number.

Solution: Let's test some prime numbers for $p$.

We have found more than five examples.

Final Answer: Five such examples are:

To Find: The number of prime and composite numbers between 21 and 30.

Solution: The numbers from 21 to 30 are: 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Let's check each number:

The prime numbers are 23 and 29. There are 2 prime numbers. The composite numbers are 21, 22, 24, 25, 26, 27, 28, 30. There are 8 composite numbers.

Given: A pair of numbers is 'safe' if Jumpy cannot reach both, meaning they have no common factor other than 1 (they are co-prime).

Solution: We check for common factors other than 1.

Final Answer: The safe pairs are (b) 4 and 15, and (c) 18 and 29.

Given: Two numbers are co-prime if their only common factor is 1.

Final Answer: The pairs of co-prime numbers are (a) 18 and 35, (b) 15 and 37, and (d) 17 and 69.

Solution: The first common multiple of two numbers is their Least Common Multiple (LCM).

1. First common multiple (LCM) is the product of the two numbers. This happens when the two numbers are co-prime (their greatest common divisor is 1).

2. First common multiple (LCM) is less than the product of the two numbers. This happens when the two numbers are not co-prime (they have a common factor greater than 1).

Relationship to co-prime numbers: The relationship is as follows:

Final Answer: $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $104 = 2 \times 2 \times 2 \times 13$ $105 = 3 \times 5 \times 7$ $243 = 3 \times 3 \times 3 \times 3 \times 3$ $320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$ $141 = 3 \times 47$ $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$ $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$ $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $1331 = 11 \times 11 \times 11$ $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

Given: The prime factors of a number are 2, 3, 3, 11.

Solution: To find the number, we multiply its prime factors. Number = $2 \times 3 \times 3 \times 11$ = $2 \times 9 \times 11$ = $18 \times 11$ = 198

Final Answer: The number is 198.

Given: The product of three prime numbers is 1955. All three primes are less than 30.

Solution: Let the number be $N = 1955$. Since the number ends in 5, one of its prime factors must be 5. 5 is a prime number less than 30. Let's divide 1955 by 5: $\frac{1955}{5} = 391$ Now we need to find two prime factors of 391, both less than 30. Let's try dividing 391 by prime numbers:

Final Answer: The three prime numbers are 5, 17, and 23.

Solution: We find the prime factorization of each factor separately and then combine them.

Final Answer: a. $56 \times 25 = 2 \times 2 \times 2 \times 7 \times 5 \times 5$ b. $108 \times 75 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5$ c. $1000 \times 81 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$

Solution: To find the smallest number with a given number of different prime factors, we should use the smallest possible prime numbers.

a. three different prime numbers The three smallest prime numbers are 2, 3, and 5. The smallest number is their product. Smallest number = $2 \times 3 \times 5 = 30$.

b. four different prime numbers The four smallest prime numbers are 2, 3, 5, and 7. The smallest number is their product. Smallest number = $2 \times 3 \times 5 \times 7 = 210$.

Final Answer: a. 30 b. 210

Solution: Two numbers are co-prime if they have no common prime factors.

Final Answer: a. No b. Yes c. No d. Yes

Solution: The first number is divisible by the second if the prime factorization of the second number is completely contained within the prime factorization of the first number.

Final Answer: a. No b. Yes c. No d. No

Are they co-prime? Two numbers are co-prime if they have no common prime factors. Here, the prime factors of the first number are {2, 3, 7} and for the second are {3, 7, 11}. They have common prime factors 3 and 7. Therefore, they are not co-prime.

Final Answer: No, they are not co-prime. No, one of them does not divide the other.

Answer: Yes, Guna is right, with the understanding that the two prime numbers are different.

Reason: Let the two different prime numbers be $p$ and $q$.

Given: Leap year rule: Divisible by 4, unless it's a century year (like 1900, 2100) not divisible by 400.

a. This question depends on the student's birth year. For example, if a student was born in 2010, the leap years until 2024 would be 2012, 2016, 2020, and 2024.

b. From the year 2024 till 2099, how many leap years are there? We need to find the number of multiples of 4 in the range [2024, 2099]. None of these years are century years, so the exception rule does not apply. The leap years are 2024, 2028, 2032, ..., 2096. This is an arithmetic sequence with first term $a_1 = 2024$, last term $a_n = 2096$, and common difference $d = 4$. Using the formula $a_n = a_1 + (n-1)d$: $2096 = 2024 + (n-1)4$ $2096 - 2024 = (n-1)4$ $72 = (n-1)4$ $\frac{72}{4} = n-1$ $18 = n-1$ $n = 19$ There are 19 leap years in this period.

Final Answer: b. From the year 2024 till 2099, there are 19 leap years.

To Find: The largest and smallest 4-digit palindromes that are divisible by 4.

Solution: A 4-digit palindrome has the form abba, where a is from 1 to 9, and b is from 0 to 9. A number is divisible by 4 if the number formed by its last two digits is divisible by 4. For the number abba, the last two digits form the number ba, which represents the value $10b + a$. So, we need $10b + a$ to be a multiple of 4.

Smallest 4-digit palindrome: We want to find the smallest abba, so we start with the smallest possible a, which is 1. If $a=1$, the number is 1bb1. The last two digits form b1. $10b+1$ must be divisible by 4. Since $10b$ is even, $10b+1$ is always odd and cannot be divisible by 4. So a cannot be 1. Let's try $a=2$. The number is 2bb2. The last two digits form b2. $10b+2$ must be divisible by 4. Let's test values for b starting from 0:

Largest 4-digit palindrome: We want to find the largest abba, so we start with the largest possible a, which is 9. If $a=9$, the number is 9bb9. The last two digits form b9. $10b+9$ is always odd, so it cannot be divisible by 4. So a cannot be 9. Let's try $a=8$. The number is 8bb8. The last two digits form b8. $10b+8$ must be divisible by 4. Let's test values for b starting from the largest, 9:

Final Answer: The smallest 4-digit palindrome divisible by 4 is 2112. The largest 4-digit palindrome divisible by 4 is 8888.

a. Sum of two even numbers gives a multiple of 4. Answer: Sometimes true. Reasoning:

b. Sum of two odd numbers gives a multiple of 4. Answer: Sometimes true. Reasoning:

Solution: We find the remainder for each number when divided by 10, 5, and 2.

Here is the table of results:

Given: The number 14560 needs to be checked for divisibility by 2, 4, 5, 8, and 10.

Solution: We need to find a pair of numbers from the list {2, 4, 5, 8, 10} such that divisibility by this pair implies divisibility by all the others. Let's analyze the relationships:

Consider the pair (8, 10):

Consider the pair (8, 5):

Let's check 14560:

Final Answer: The two numbers could be 8 and 5 (or 8 and 10).

Solution: A number is divisible by all of 2, 4, 5, 8, and 10 if it is divisible by their Least Common Multiple (LCM). LCM(2, 4, 5, 8, 10) = LCM(LCM(8, 10), 2, 4, 5) = LCM(40, 2, 4, 5) = 40. So, we need to check which of the given numbers are divisible by 40. A number is divisible by 40 if it ends in 0 and the number formed by the remaining digits is divisible by 4.

Final Answer: The numbers divisible by all of 2, 4, 5, 8 and 10 are 5600, 6000, and 77622160.

Given: The product of two numbers is 10000. Neither number should end in 0.

Solution: A number ends in 0 if it is a multiple of 10, which means it has both 2 and 5 as prime factors. First, let's find the prime factorization of 10000. $10000 = 100 \times 100 = (10 \times 10) \times (10 \times 10) = 10^4 = (2 \times 5)^4 = 2^4 \times 5^4$.

We need to split these prime factors ($2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5$) into two groups to form two numbers. To ensure that neither number ends in 0, we must not give both a factor of 2 and a factor of 5 to the same number. Therefore, one number must be formed using only the factors of 2, and the other number must be formed using only the factors of 5.

Let's check:

Final Answer: The two numbers are 16 and 625.

Puzzle 1 Rows: 105, 20, 30. Cols: 28, 125, 18.

Puzzle 2 Rows: 8, 105, 70. Cols: 30, 70, 28.

Puzzle 3 Rows: 63, 27, 190. Cols: 45, 42, 171.

Puzzle 4 Rows: 343, 66, 44. Cols: 28, 154, 231.