Joe Ganley

My daughter brought home a fun math problem last week; it took a little mental effort to solve, and I haven't found a solution on the web, so I thought I'd post one.

The problem is this: What is the largest n such that 1005! is evenly divisible by 10ⁿ? Put another way, how many 0's does the expansion of 1005! end with?

Highlight the following blank space for the solution:

Imagine writing out the expansion of 1005!, and then replacing each number with its prime factorization. The prime factors of 10 are 5 and 2, so our question is equivalent to asking how many (5,2) pairs this expansion contains. It is easy to see that it contains far more 2's than 5's, so the question becomes: How many 5's are in the expansion? Every number in the original expansion of 1005! that is divisible by 5 contributes a 5 in the prime-factorization expansion; there are 201 of these. However, every number that is divisible by 5², i.e. 25, contributes a second 5; this adds 40 more. Every number that is divisible by 5³, i.e. 125, contributes a third 5; add 8 more. Finally, the one number that is divisible by 5⁴, i.e. 625, contributes a fourth 5, adding one more and bringing the total count to n = 250.

Labels: math, puzzles

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Joe Ganley	I make software and sometimes other things.
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1005!
My daughter brought home a fun math problem last week; it took a little mental effort to solve, and I haven't found a solution on the web, so I thought I'd post one. The problem is this: What is the largest n such that 1005! is evenly divisible by 10ⁿ? Put another way, how many 0's does the expansion of 1005! end with? Highlight the following blank space for the solution: Imagine writing out the expansion of 1005!, and then replacing each number with its prime factorization. The prime factors of 10 are 5 and 2, so our question is equivalent to asking how many (5,2) pairs this expansion contains. It is easy to see that it contains far more 2's than 5's, so the question becomes: How many 5's are in the expansion? Every number in the original expansion of 1005! that is divisible by 5 contributes a 5 in the prime-factorization expansion; there are 201 of these. However, every number that is divisible by 5², i.e. 25, contributes a second 5; this adds 40 more. Every number that is divisible by 5³, i.e. 125, contributes a third 5; add 8 more. Finally, the one number that is divisible by 5⁴, i.e. 625, contributes a fourth 5, adding one more and bringing the total count to n = 250. Labels: math, puzzles Comments (0)