Difference between revisions of "2017 AMC 8 Problems/Problem 12"
Nukelauncher (talk | contribs) m (→Solution) |
Topnotchmath (talk | contribs) |
||
| Line 9: | Line 9: | ||
==See Also== | ==See Also== | ||
| − | {{AMC8 box|year=2017|num-b=11| | + | {{AMC8 box|year=2017|num-b=11|num-a=13}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:13, 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 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. 
