Reversing the DigitsDate: 05/29/2003 at 19:00:19 From: Sarah Subject: Reversing digits Certain pairs of numbers that have two digits have the same product when you reverse their digits. For example: 12 x 42 = 21 x 24 504 504 24 x 63 = 42 x 36 1512 1512 ...etc My teacher says there are 11 other possibilities. Is there a pattern to follow to find them, or should I randomly check every number until it works, as I have been doing up until now? Date: 05/29/2003 at 23:21:45 From: Doctor Ian Subject: Re: Reversing digits Hi Sarah, I don't know if there's a formula, but maybe we can be a little more clever about finding the answers. Suppose our numbers look like 'ab' and 'cd'. Then we can represent them as 10a + b and 10c + d Do you see why? Using a representation like this is often the key to solving this kind of problem. We're going to multiply the numbers to get (10a + b)(10c + d) and we're going to multiply them with the digits switched, (10b + a)(10d + c) and the products are supposed to be equal: (10a + b)(10c + d) = (10b + a)(10d + c) If we expand this, we get 100ac + 10bc + 10ad + bd = 100bd + 10ad + 10bc + ac Now, that looks like a mess, but if we look closely, we can see that 10bc and 10ad appear on both sides of the equation. So they aren't really contributing anything. Let's get rid of them: 100ac + bd = 100bd + ac This is looking more interesting. Let's subtract bd from both sides, and ac from both sides: 100ac - ac = 100bd - bd 99ac = 99bd ac = bd So what we're _really_ looking for are pairs of single-digit factors that give the same product. For example, a c b d 24 = 3*8 = 6*4 So we would expect 10*3 + 6 = 36 and 10*8 + 4 = 84 to be one of the pairs we're looking for. Does it work? 36 * 84 = 63 * 48 3024 = 3024 Excellent! Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/