Difference between revisions of "1951 AHSME Problems/Problem 37"
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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> throught <math>10</math> is <math>2520</math>, therefore the number we want to find is <math>2520-1=\boxed{\textbf{(D)}\ 2519}</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> throught <math>10</math> is <math>2520</math>, therefore the number we want to find is <math>2520-1=\boxed{\textbf{(D)}\ 2519}</math> | ||
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| + | == See Also == | ||
| + | {{AHSME 50p box|year=1951|num-b=33|num-a=35}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 12:46, 2 February 2013
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 throught is 
, therefore the number we want to find is 
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
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| All AHSME Problems and Solutions | ||
