Puzzles Based on Divisibility Rules

Let's play a number game!

Here's how it works:

(1) Think of any two-digit number (but don't tell me what it is yet)

(2) Reverse the digits of your number to get a new number

(3) Add these two numbers together

(4) Divide your result by 9

I bet the remainder will always be 0!

Let me explain why this works, similar to the original puzzle:

Let's say you choose a two-digit number represented as (10a + b), where 'a' is the digit (1-9) and 'b' is the digit (0-9)

When you reverse it, you get (10b + a)

Adding them together: (10a + b) + (10b + a) = a + b = (a + b)

Since the result is always 11(a + b), it will always be divisible by 11 in the original puzzle.

In this variation, the result 11(a + b) will always be divisible by 9 because the sum of digits of any number divisible by 9 is also divisible by 9.

We can verify this: If someone picks 75:

Original:

Reversed:

Sum:

132 ÷ 9 = where remainder is .

Chapter 15: Playing with Numbers