Key Points

To understand 'Think of a Number' tricks, represent the unknown number with a variable like $x$. Apply the given arithmetic operations to $x$ to form an algebraic expression. Simplifying this expression often reveals a constant value, explaining why the trick works regardless of the starting number.

A date with month $M$ and day $D$ can be found from a final answer. The trick creates an expression like $100M + D + k$. To find the date, subtract the constant $k$ from the final number. The last two digits of the result give the day $D$, and the preceding digits give the month $M$.

In a number pyramid, the value in any box is the sum of the values in the two boxes directly below it. This rule allows you to set up and solve linear equations to find unknown numbers within the pyramid.

For a pyramid with three rows, if the bottom row consists of numbers $a, b, c$ from left to right, the number at the very top is given by the expression $a + 2b + c$.

For a pyramid with four rows, if the bottom row consists of numbers $a, b, c, d$, the number at the top is given by the expression $a + 3b + 3c + d$. The coefficients follow the pattern of Pascal's triangle.

Numbers in a calendar grid have fixed relationships. In a $2 imes 2$ grid, if the top-left number is $a$, the other numbers are $a+1$, $a+7$, and $a+8$. The sum of these four numbers is $4a + 16$, which can be used to find all numbers if the sum is known.

To get the largest product from three digits $p, q, r$ with $p < q < r$ in the form $□□ imes □$, place the largest digit $r$ as the single multiplier. Arrange the other two digits in descending order to form the two-digit number. The expression for the largest product is $(10q + p) imes r$.

The difference between any two-digit number $10a+b$ and its reverse $10b+a$ is always divisible by 9. The difference is either $9(a-b)$ or $9(b-a)$.

The sum of any two-digit number $10a+b$ and its reverse $10b+a$ is always divisible by 11. The sum is $11(a+b)$.

The sum of a 3-digit number $abc$ and its cyclic permutations $bca$ and $cab$ is always divisible by 3, 37, and 111. The sum is $111(a+b+c)$, and $111 = 3 imes 37$.

A 6-digit number formed by repeating a 3-digit number, like $abcabc$, is always divisible by 7, 11, and 13. This is because $abcabc = abc imes 1001$, and $1001 = 7 imes 11 imes 13$.

To solve word problems using algebra, first identify the unknown quantity and represent it with a variable. Then, translate the problem's statements and conditions into a linear equation. Solving this equation gives the value of the unknown.

For problems involving a sequence of operations, it is often effective to work backwards from the final result. Reverse each operation to find the initial value. For example, to reverse 'double and then subtract 8', you must 'add 8 and then divide by 2'.