Difference between revisions of "2014 AMC 8 Problems/Problem 8"
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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> | 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 | ~fn106068 | ||
| + | |||
| + | ==Video Solution== | ||
| + | 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}} | ||
Revision as of 08:15, 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. 
