Difference between revisions of "2017 AMC 8 Problems/Problem 12"
(→Solution) |
Nukelauncher (talk | contribs) m (→Solution) |
||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
− | Finding the LCM of 4, 5, and 6; we find | + | Finding the LCM of <math>4</math>, <math>5</math>, and <math>6</math>; we find it is <math>60</math>. Now, <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 15:43, 22 November 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
Finding the LCM of , , and ; we find it is . Now, , and that is in the range of
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by 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.