Difference between revisions of "2014 AMC 8 Problems/Problem 8"
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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>\textbf{(D) }3</math>. | ||
+ | ~SparklyFlowers | ||
+ | |||
+ | ==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> | ||
+ | ~fn106068 | ||
+ | |||
+ | ==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 09:15, 27 April 2022
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 numbers added together minus the even numbers added together(or vice versa) is a multiple of . So, we have = a multiple of . The only multiple that works here is , as . Thus, ~fn106068
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.