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

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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4</math>
 
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4</math>
  
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==Solution==
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==Solution 1==
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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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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>
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~SparklyFlowers
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==Solution 2==
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We know that a number is divisible by <math>11</math> if the odd digits added together minus the even digits added together (or vice versa) is a multiple of <math>11</math>. Thus, we have <math>1+2-A</math> = a multiple of <math>11</math>. The only multiple that works here is <math>0</math>, as <math>11 \cdot 0 = 0</math>. Thus, <math>A = \boxed{\textbf{(D)}~3}</math>
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~fn106068
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/kNHe_IMcMiU
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/mHWTWk-xt0o ~savannahsolver
  
 
==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}}

Latest revision as of 10:14, 2 July 2023

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 digits added together minus the even digits added together (or vice versa) is a multiple of $11$. Thus, we have $1+2-A$ = a multiple of $11$. The only multiple that works here is $0$, as $11 \cdot 0 = 0$. Thus, $A = \boxed{\textbf{(D)}~3}$ ~fn106068

Video Solution (CREATIVE THINKING)

https://youtu.be/kNHe_IMcMiU

~Education, the Study of Everything


Video Solution

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

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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