problem of the week

mathisfun

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Sep 10, 2005
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Find the smallest positive integer that is divisible by 13 and that when divided by any of the integers from 2 to 12 (including both) leaves a remainder of 1.
 
problem

First, i found the factors of 2 through 12.
The smallest number dividible by all of them is 2^3*3^2*5*7*11 = 277720. Now, i divide that number by 13 to find the remainder. I get a remainder of 1. Now, to find number divisible by 13, i know it has to be a multiple of this number plus 1.
Now, since i found that it has a remainder of 1 after dividing by 13,
I realize that if i multiply it by 12, then it will give me a number divisible by 13 plus 12. And since i add 1 to that then it would give me a number divisible by 13 plus 12+1, or, 13.
So the number i get is 12*277720 +1, or, 3332641.
 
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