Lizzie came up with a divisibility test for a certain number m is not equal to 1:

Break a positive integer n into two-digit chunks, starting from the ones place. (For example, the number 354764 would break into the two-digit chunks 64,35,47)

Find the alternating sum of these two-digit numbers, by adding the first number, subtracting the second, adding the third, and so on. (In our example, this alternating sum would be 64-47+35=52 )

Find and show that this is indeed a divisibility test for (by showing that n is divisible by m if and only if the result of this process is divisible by m).

Sunghoon Aug 19, 2023

#1**0 **

Let m=37. We can show that n is divisible by 37 if and only if the result of this process is divisible by 37 as follows:

If n is divisible by 37, then the alternating sum of the two-digit chunks is divisible by 37.

This is because the alternating sum of the two-digit chunks is just the sum of the digits of n, modulo 37. Since n is divisible by 37, the sum of its digits is also divisible by 37, so the alternating sum of the two-digit chunks is also divisible by 37.

If the alternating sum of the two-digit chunks is divisible by 37, then n is divisible by 37.

This is because the sum of the digits of n is congruent to the alternating sum of the two-digit chunks, modulo 37. Therefore, if the alternating sum of the two-digit chunks is divisible by 37, then the sum of the digits of n is also divisible by 37, so n is divisible by 37.

Therefore, the alternating sum of the two-digit chunks is a divisibility test for 37.

To see this in action, let's consider the number 354764. The alternating sum of the two-digit chunks of 354764 is 64−47+35=52. Since 52 is divisible by 37, we know that 354764 is also divisible by 37.

Guest Aug 19, 2023