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.
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
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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