Consecutive Integers, Divisible by the Sums of Their Digits
Date: 05/20/2002 at 05:20:20 From: Mike Subject: Problem Solving Find the first three consecutive integers greater than 10, each of which is divisible by the sum of its digits. I know that the numbers are 110, 111, and 112. But I don't know how to explain why they are the first numbers that work. Could you please help me out? Thank you.
Date: 05/20/2002 at 13:07:59 From: Doctor Mike Subject: Re: Problem Solving Hi, It's always nice to hear from another Mike. Another word for "explain" that mathematicians use for what you are talking about is "prove". A proof is a logical sequence of steps that will demonstrate, beyond any doubt, that the given answer is true. This idea of proof is the heart of mathematics. It's good you are getting started on it. One method of proof, which may be the best choice here, is called the "brute force" technique. No fancy logic needed; only straightforward observation. You can just check every number from 11 to 112 and write down in a list which ones are actually divisible by the sum of their digits. The list will start out 12, 18, 20, 21, 24, 27, 30 and so on. When you get to the point of adding 112 to the list, look back on the list and observe that there were not any 3 in a row until that point. Done! Of course there is a possibility you made some kind of mistake, so you should check your work over for such things. (Even experienced mathematicians make mistakes, but they are smart enough to know that about themselves, and always re-check the work they have done to make sure they have not missed anything important.) After you have done this, and in fact ANY time you prove something, it is good to try to find a better way. You may find out something that allows you to explain why it is true, but in a way that is more quick or clever. I did this but did not come up with one here. However, I did notice some interesting patterns in the list. I saw that ANY 2-digit number which ends in the digit zero is automatically on the list. Why? Because such a number "x0" has x+0=x as the sum of its digits, and the number equals x times 10 so the number is divisible by x. What other patterns are there? When I made my list I did it by writing the numbers down in rows and columns, with the teens first, then the 20s in a row, then the 30s, etc. I circled the ones for the list, and marked out with an X all the rest. Geometrically, I noticed that there was sort of a "diagonal" which had circles around them. On closer inspection I saw that these were the 2-digit numbers whose digits summed to nine. That is 18, 27, 36, 45, 54, 63, 72, 81 and 90. "Gee, why is that?" I asked myself. Here is an opportunity for using some real logic (and algebra) to prove something. Any such number is of the form "xy" where x+y = 9. Is xy divisible by 9? Yes, because xy = 10*x + y = 10*x + (9-x) = 10*x - x + 9 = 9x + 9 and 9x+9 factors into 9*(x+1). Pretty slick, right? You should go thru and tell why each "=" step is correct. It's not hard or tricky at all. That is about the extent of the regularity that I saw, so the most efficient and clear way to explain it seems to be the way I tried first. I hope this has helped. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/
Date: 05/22/2002 at 04:53:32 From: Mike Subject: Problem Solving Hi Doc Mike, Thank you very much for your help with the math problem, it was very much appreciated. Your information really helped me out. I have some other problems which are a bit trickier. Maybe I will send them to you some time. Thanks again, Mike
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