Date: 11/15/2001 at 22:51:01 From: Jannet Stockton Subject: Math 1) How many nonnegative integers consisting of one to three digits are divisible by 5? Leading zeros are not allowed. I don't know that it includes 0 or not; if it includes 0, the answer is 200. 2) How many nonnegative integers consisting of one to three different digits are divisible by 5? Leading zeros are not allowed. The answer is 155, but how can I get it?
Date: 11/16/2001 at 16:30:16 From: Doctor Ian Subject: Re: Math Hi Jannet, Zero is non-negative, so I would say it's included. For number 2, since you already know how many integers are divisible by 5, all you have to do is subtract the ones that have duplicate digits, e.g., 110, 225, 505, and so on. Apparently there are 45 such numbers, so you need to show that there is a systematic way to find them... i.e., one that is guaranteed to get all of them. You could start by thinking about the patterns that the numbers would have to fall into to (1) be divisible by 5 and (2) have a duplicate digit. Here are some: aa0 There would be 9 like this. aa5 And 9 more like this. 5a5 And 9 more like this. a00 And 9 more like this. That gives you 36. If you can find one more pattern, you'll have found them all. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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