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 N-Divisibility (Posted on 2004-02-29)
How many positive integers divide at least one of 10^40 and 20^30?

 See The Solution Submitted by DJ Rating: 4.1111 (9 votes)

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 solution | Comment 2 of 4 |

10^40 = 2^40 * 5^40
20^30 = 2^60 * 5^30

The divisors of 10^40 can, therefore, have zero through 40 factors equal to 2 and zero through 40 factors equal to 5. Combining these choices, there are 41*41=1681 divisors of the first number.

The second number, similarly, can have 2 appear as a factor zero to 60 times and 5 as a factor zero to 30 times, for 61*31 = 1891 divisors.

But some of these 1681+1891 numbers are divisors of both numbers and therefore have been counted twice.  Those are the numbers that have 2 as a factor zero to 40 times and 5 as a factor zero to 30 times, accounting for 41*31=1271 divisors, so we get 1681+1891-1271 = 2301 divisors.

 Posted by Charlie on 2004-02-29 11:24:09

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