Art of Problem Solving
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2014 AMC 8 Problems/Problem 8: Difference between revisions

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==Solution==
==Solution==
A number is divisible by <math>11</math> if the difference between the sum of the digits in the odd-numbered slots (e.g. the ones slot, the hundreds slot, etc.) and the sum in the even-numbered slots (e.g. the tens slot, the thousands slot) is a multiple of <math>11</math>. So <math>1 + 2 - A</math> is equivalent to <math>0\pmod{11}</math>. Clearly <math>1+2-A</math> cannot be equal to <math>11</math> or any multiple of <math>11</math> greater than that. Also, if the expression <math>1+2-A</math> is to be equal to a negative multiple of <math>A</math>, <math>A</math> must be 14 or greater, which violates the condition that A is a digit. So <math> 3 - A = 0 \longrightarrow A = \boxed{\textbf{(D)}~3}</math>.
The only 3-digit number that is divisible by 11 and has a units digit of 2 is 132 so A is \boxed{\textbf{(D)}~3}$.


==See Also==
==See Also==
{{AMC8 box|year=2014|num-b=7|num-a=9}}
{{AMC8 box|year=2014|num-b=7|num-a=9}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 11:49, 27 November 2020

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

The only 3-digit number that is divisible by 11 and has a units digit of 2 is 132 so A is \boxed{\textbf{(D)}~3}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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