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

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==Solution==
 
==Solution==
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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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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~mod~11</math>. Clearly 3 , (A) cannot be equal to <math>11</math> or any multiple of <math>11</math> greater than that. So <math> 3 - A = 0 \longrightarrow A = 3</math>.
  
 
==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 10:22, 28 November 2014

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

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 $3 - A = 0 \longrightarrow A = 3$.

See Also

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

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