Difference between revisions of "2014 AMC 8 Problems/Problem 8"
(rewrote sol because it had some issues (not explaining)) |
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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> | ||
| − | ==Solution== | + | ==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>. | 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 | ~SparklyFlowers | ||
| + | |||
| + | ==Solution 2== | ||
| + | We know that a number is divisible by 11 if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of 11. So, we have <math>1+2-A</math> = a multiple of ll. The only multiple that works here is 0, as 11x0 = 0. Thus, <math>A = \qquad\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 13:31, 10 June 2021
Contents
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 11 if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of 11. So, we have 
= a multiple of ll. The only multiple that works here is 0, as 11x0 = 0. Thus, 
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. 
