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> |
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| + | ==Video Solution== | ||
| + | https://youtu.be/UQGamWE1XOo | ||
| + | |||
| + | ~Lucas | ||
== See Also == | == See Also == | ||
| − | {{AHSME 50p box|year=1951|num-b= | + | {{AHSME 50p box|year=1951|num-b=36|num-a=38}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:27, 19 September 2022
Contents
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 
Video Solution
~Lucas
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
