Difference between revisions of "2014 AMC 8 Problems/Problem 8"

Revision as of 01:59, 24 December 2022

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 $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

Solution 1

Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{\textbf{(D)}~3}$ ~SparklyFlowers

Solution 2

We know that a number is divisible by $11$ if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of $11$ . So, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0 = 0$ . Thus, $A = \textbf{(D) }3$ ~fn106068

Video Solution

https://youtu.be/mHWTWk-xt0o ~savannahsolver

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Since all the eleven members paid the same amount, that means that the total must be divisible by <math>11</math>. We can do some trial-and-error to get <math>A=3</math>, so our answer is <math>\textbf{(D) }3</math>.
+Since all the eleven members paid the same amount, that means that the total must be divisible by <math>11</math>. We can do some trial-and-error to get <math>A=3</math>, so our answer is <math>\boxed{\textbf{(D)}~3}</math>
 ~SparklyFlowers

Page

Toolbox

Search