Difference between revisions of "1951 AHSME Problems/Problem 37"
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==Solution== | ==Solution== | ||
| − | If we add <math>1</math> to the number, it becomes divisible by <math>10, 9, 8, \cdots, 2, 1</math>. The LCM of <math>1</math> | + | If we add <math>1</math> to the number, it becomes divisible by <math>10, 9, 8, \cdots, 2, 1</math>. The LCM of <math>1</math> through <math>10</math> is <math>2520</math>, therefore the number we want to find is <math>2520-1=\boxed{\textbf{(D)}\ 2519}</math> |
== See Also == | == See Also == | ||
Revision as of 13:11, 16 October 2016
Problem 37
A number which when divided by 
leaves a remainder of 
, when divided by leaves a remainder of 
, by leaves a remainder of 
, etc., down to where, when divided by 
, it leaves a remainder of 
, is:
Solution
If we add to the number, it becomes divisible by 
. The LCM of through is 
, therefore the number we want to find is 
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
