1005! / 10^n: A Number PuzzleDate: 9/11/96 at 17:38:17 From: Anonymous Subject: 1005! / 10^n: A Number Puzzle Find the largest positive integer, n, such that 1005! is divisible by 10 to the power of n. I would like to know: a) how one would go about solving such a problem and b) your opinion of the grade level of the problem Date: 9/12/96 at 14:39:6 From: Doctor Ceeks Subject: Re: 1005! / 10^n: A Number Puzzle Hi, a) The problem is about prime factorization. Given any number, it can be factored as 2^X 5^Y M, where M has no factors of 2 or 5. Since 10 = 2*5, the number of tens which divide the number is the minimum of X and Y...each time you divide by 10, you eliminate a factor of 2 and a factor of 5, and you can keep dividing until you run out of 2's or 5's (you have to have both). So, we have to figure out how many factors of 2 and how many factors of 5 there are in 1005!=1005*1004*1003*...*3*2*1 1005/5 = 201 of the numbers in the above product are divisible by 5. [1005/25]=40 are divisible by 25. ([x]=greatest integer < or = to x) [1005/125]=8 are divisible by 125=5^3. [1005/625]=1 is divisible by 625=5^4. [1005/3025]=0 are divisible by 3025=5^5. We have to make sure we count the number of factors of five properly. 201 contribute at least 1 five, and of those, 40 give at least 2, 8 at least 3, and 1 gives 4. So the total number of factors of five in 1005! is 201+40+8+1=250. You do the same computation for factors of 2, and get some number. It will be bigger than 250. So the answer is: n = 250. Another way of saying it is that 1005! is a number ending with 250 zeroes. b) I cannot answer which grade level... it's hard to answer that one. Different people learn different things at different times. So if grade level is supposed to measure what most kids are supposed to be able to do by a certain age, then I don't have the data for this particular problem. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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