Difference between revisions of "2005 iTest Problems/Problem 25"
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If for any <math>N</math>, <math>N!</math> is divisible by 2005, any integers greater than <math>N</math> also share that property. So we should start by finding the smallest factorial that is divisible by 2005. Since <math>2005=5\cdot401</math>, and 401 is prime, the smallest such <math>N</math> is 401, as no smaller factorials are divisible by 401. Thus there are <math>401-1=400</math> factorials that are not divisible by 2005, and our answer is <math>2005-400=\boxed{1605}</math> such elements. | If for any <math>N</math>, <math>N!</math> is divisible by 2005, any integers greater than <math>N</math> also share that property. So we should start by finding the smallest factorial that is divisible by 2005. Since <math>2005=5\cdot401</math>, and 401 is prime, the smallest such <math>N</math> is 401, as no smaller factorials are divisible by 401. Thus there are <math>401-1=400</math> factorials that are not divisible by 2005, and our answer is <math>2005-400=\boxed{1605}</math> such elements. | ||
==See Also== | ==See Also== | ||
| − | {{iTest box|year=2005|num-b=24|num-a= | + | {{iTest box|year=2005|num-b=24|num-a=26}} |
[[Category: Introductory Number Theory Problems]] | [[Category: Introductory Number Theory Problems]] | ||
Latest revision as of 19:31, 13 October 2025
Problem
Consider the set 
. How many elements of this set are divisible by 
?
Solution 1
If for any 
, 
is divisible by 2005, any integers greater than also share that property. So we should start by finding the smallest factorial that is divisible by 2005. Since 
, and 401 is prime, the smallest such is 401, as no smaller factorials are divisible by 401. Thus there are 
factorials that are not divisible by 2005, and our answer is 
such elements.
See Also
| 2005 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 24 |
Followed by: Problem 26 | |
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