## Test of Divisibility by 11

According to the test of divisibility **rule** of **11**, we must subtract and then add the digits in an alternating pattern from left to right. If the answer is 0 or **11**, then the result is divisible by **11**.

## Try the problem

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker\(1A_2\) . What is the missing digit A of this 3 -digit number?

AMC 8

Number System

5 out of 10

Challenges and Thrills in Pre-College Mathematics

Excursion of Mathematics

## Knowledge Graph

## Use some hints

Given, eleven members of the Middle School Math Club each paid the same amount for a guest speaker. So the total amount paid to the guest must be divisible by 11 .Therefore , 1A2 is divisible by 11.

Now remind the test of divisibility by 11.So , 1+2- A is divisible by 11.

Clearly , 1+2 – A cannot be equal to 11 or any multiple of 11 greater than that as A is a digit and it lies between 0 to 9. Also, if the expression 1+2 – A is equal to a negative multiple of A. A must be 14 or greater, which violated the condition that A is a digit.

So, 3-A=0 and 0 is divisible by any number. Hence, A=3.

## Other Useful links

- https://www.cheenta.com/perimeter-of-a-circle-amc-8-2013-problem-25/
- https://www.youtube.com/watch?v=6ClB57eYnbo