2014 AMC 8 Problems/Problem 8: Difference between revisions
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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 | ==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 <math> \textdollar\underline{1} \underline{A} \underline{2}</math>. What is the missing digit <math>A</math> of this <math>3</math>-digit number? | |||
<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> | ||
==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>\boxed{\textbf{(D)}~3}</math> | |||
~SparklyFlowers | |||
==Solution 2== | |||
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 | |||
==Solution 3== | |||
If you know your multiplication tables up to <math>12*12</math>, you know that the out of the choices provided, the only number with <math>1A2</math> that is divisible by <math>11</math> is <math>132</math> or <math>12*11</math>. So the answer is <math>\boxed{\textbf{(D)}~3}</math>. ~Jupiter10 | |||
==Video Solution (CREATIVE THINKING)== | |||
https://youtu.be/kNHe_IMcMiU | |||
~Education, the Study of Everything | |||
==Video Solution== | |||
https://youtu.be/mHWTWk-xt0o ~savannahsolver | |||
==See Also== | |||
{{AMC8 box|year=2014|num-b=7|num-a=9}} | |||
{{MAA Notice}} | |||
Latest revision as of 23:21, 17 January 2025
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 
. What is the missing digit 
of this 
-digit number?
Solution 1
Since all the eleven members paid the same amount, that means that the total must be divisible by 
. We can do some trial-and-error to get 
, so our answer is 
~SparklyFlowers
Solution 2
We know that a number is divisible by if the odd digits added together minus the even digits added together (or vice versa) is a multiple of . Thus, we have 
= a multiple of . The only multiple that works here is 
, as 
. Thus, 
~fn106068
Solution 3
If you know your multiplication tables up to 
, you know that the out of the choices provided, the only number with 
that is divisible by is 
or 
. So the answer is . ~Jupiter10
Video Solution (CREATIVE THINKING)
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
