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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 <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 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 = \boxed{\textbf{(D)}~3}</math>.
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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\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>.
  
 
==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 18:21, 13 November 2017

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 $0\pmod{11}$. Clearly $1+2-A$ cannot be equal to $11$ or any multiple of $11$ greater than that. Also, if the expression $1+2-A$ is to be equal to a negative multiple of $A$, $A$ must be 14 or greater, which violates the condition that A is a digit. So $3 - A = 0 \longrightarrow A = \boxed{\textbf{(D)}~3}$.

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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