Difference between revisions of "2017 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | + | The LCM(4,5,6) is 60. Since <math>60+1=61</math>, and that is in the range of <math>\boxed{\textbf{(D)}\ \text{60 and 79}}.</math> | |
==See Also== | ==See Also== | ||
Revision as of 18:10, 11 December 2017
Problem 12
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?
Solution
The LCM(4,5,6) is 60. Since 
, and that is in the range of 
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
