Difference between revisions of "2014 AMC 8 Problems/Problem 8"
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==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}} |
Latest 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.