Dividing 1005! by 10^NDate: 9/5/96 at 16:33:24 From: Anonymous Subject: Largest Power of 10 that Divides... Determine the largest interger N such that 1005! is divisable by 10^N. Get the largest power of 10 that will divide into 1005! I multiplied the first three numbers of the series and saw a pattern that 3 place values increased with each operation. My question is, how do I determine the equation to solve such a problem? My calculator can't handle the big numbers and I don't have a supercomputer! Sincerely, Colleen Date: 9/6/96 at 20:7:59 From: Doctor Tom Subject: Re: Largest Power of 10 that Divides... Hi Colleen, Here's the way I'd think about it. Each 10 that divides the giant product means that there was a factor of 5 and a factor of 2. Since the factors of two are much more common than the factors of 5, there will be plenty of factors of 2, and so we just need to count the factors of 5. Let's look at a simpler problem. How would you work it out for 21! ? Well 5 goes evenly into 5, 10, 15, and 20, so there are four 5s, so there will be four zeroes at the end of this product. How about 32! ? Well, 5, 10, 15, 20, 25, and 30 are in the product so at first, the answer looks like 6, but it's actually 7, since there are 2 factors of 5 in 25. How about 150! ? Find the number of factors of 5: 150/5 = 30 Factors of 25: 150/25 = 6 and we have to worry about factors of 125, since it has 3 factors of 5 (125 = 5*5*5), and there's one of those less than 150. So there are 30+6+1 = 37 zeroes at the end of this number. For your 1005!, you'll also have to worry about factors of 625 = 5*5*5*5, but there won't be any larger ones to worry about because 5*5*5*5*5 is bigger than 1005. I won't tell you the answer, but I think this should be a good enough hint. Try to work it with these hints, and if you can't ask again. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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