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AMC 8 USA Math Olympiad

Test of Divisibility by 11- AMC 8, 2014 – Problem-8

The simplest example of power mean inequality is the arithmetic mean – geometric mean inequality. Learn in this self-learning module for math olympiad

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


Test of Divisibility of 11- 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.



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