Difference between revisions of "2017 AMC 8 Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder | + | Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The <math>\operatorname{LCM}(4,5,6)</math> is <math>60</math>. 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 09:11, 23 August 2019
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
Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The 
is 
. 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. 
