Difference between revisions of "2014 AMC 8 Problems/Problem 8"
(→Solution) |
(rewrote sol because it had some issues (not explaining)) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | + | 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 | ||
==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 20:22, 24 January 2021
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
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
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. 
