Difference between revisions of "2012 AMC 8 Problems/Problem 15"
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<math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99 </math> | <math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99 </math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/rQUwNC0gqdg?t=172 | ||
==Solution== | ==Solution== | ||
Revision as of 19:26, 27 October 2020
Contents
Problem
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=172
Solution
To find the answer to this problem, we need to find the least common multiple of 
, 
, 
, 
and add 
to the result. The least common multiple of the four numbers is 
, and by adding , we find that that such number is 
. Now we need to find the only given range that contains . The only such range is answer 
, and so our final answer is 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
