Algebra Play - Thinking about ‘Think of a Number’ Tricks

Megha asks her friend to: 1. Think of a number. 2. Double it. 3. Add $10$. 4. Divide the result by $2$. 5. Subtract the original number. Show algebraically why the answer is always $5$.

Let the number thought by the friend be $x$. Step 1: $x$ Step 2: Double it: $2x$ Step 3: Add $10$: $2x + 10$ Step 4: Divide by $2$: $\frac{2x + 10}{2} = \frac{2(x + 5)}{2} = x + 5$ Step 5: Subtract the original number: $(x + 5) - x = 5$

By simplifying the algebraic expression, the variable $x$ is eliminated, leaving a constant result of $5$ regardless of the initial number chosen.

Take a two-digit number $72$. Reverse the digits to get $27$. Find the sum and show it is divisible by $11$.

Original number: $72$ Reversed number: $27$ Sum: $\begin{array}{r} 72 \\ +27 \\ \hline 99 \end{array}$ Dividing by $11$: $\frac{99}{11} = 9$.

Using the general form $(10a + b) + (10b + a) = 11(a + b)$, we see the sum is $11(7 + 2) = 11 \times 9 = 99$. Any such sum is a multiple of $11$.

Choose a three-digit number where the first digit is greater than the last. Let the number be $843$. Reverse it and subtract the smaller from the larger. Show the result is divisible by $99$.

Original number: $843$ Reversed number: $348$ Difference: $\begin{array}{r} 843 \\ -348 \\ \hline 495 \end{array}$ Dividing by $99$: $\frac{495}{99} = 5$.

Algebraically, $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$. Here $a=8, c=3$, so $99(8 - 3) = 99 \times 5 = 495$.