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

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==Solution 1==
 
==Solution 1==
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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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>
 
~SparklyFlowers
 
~SparklyFlowers
  
 
==Solution 2==
 
==Solution 2==
We know that a number is divisible by <math>11</math> if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of <math>11</math>. So, 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 = \textbf{(D) }3</math>
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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>
 
~fn106068
 
~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 11:14, 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 (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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