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

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==Solution==
 
==Solution==
A number is divisible by 11 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 11. So 1 + 2 - A is equivalent to 0 mod 11. Clearly 3 - A cannot be equal to 11 or any multiple of 11 greater than that. So 3 - A = 0 --> A = 3.
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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>.
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~SparklyFlowers
  
 
==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 21:22, 24 January 2021

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

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 $\textbf{(D) }3$. ~SparklyFlowers

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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