Products of DigitsDate: 03/02/2003 at 18:24:08 From: Liz Subject: Product of digits The product of digits in the number 234 is 24: 2*3*4 = 24. Can you describe a general procedure for figuring out how many x-digit numbers have a product equal to p, where x and p are counting numbers? How many different three-digit numbers have a product equal to 12? to 18? Why does it work out that way? Date: 03/03/2003 at 12:06:15 From: Doctor Ian Subject: Re: Product of digits Hi Liz, A 'general procedure' would be to start by finding the prime factors of p. For example, if p = 12, the prime factors are 12 = 2 * 2 * 3 We can use each prime factor as a digit, so now we want to know how many distinct ways we can arrange them. If we treat the 2's as individuals (2 and 2'), we get 22'3 232' 2'23 2'32 322' 32'2 But of course, they aren't distinguishable, so we end up with half as many possibilities: 223 232 322 Once we've identified the digits, we can use rules for computing permutations to generate the number of possibilities without having to write them all down. See the Dr. Math FAQ: Permutations and Combinations http://mathforum.org/dr.math/faq/faq.comb.perm.html For example, in this case, we had three digits to work with, so the number of arrangements would be n = 3! = 3 * 2 * 1 But two of the digits were the same, and those could be arranged in n = 2! = 2 * 1 different ways. So we end up with n = 3! / 2! = 6 / 2 = 3 distinct arrangements of digits. Now, this is a particularly simple case, because the number of prime factors is the same as the number of digits. But consider what happens when p = 24. Now the prime factors are 24 = 2 * 2 * 2 * 3 and we can group these in a couple of different ways: 2 * (2*2) * 3 digits 2, 4, 3 2 * 2 * (2*3) digits 2, 2, 6 So now we can proceed as before, except using (2,3,4) and (2,2,6) as if they were prime factors. In general, any time you're dealing with prime factors, it's hard to come up with simple formulas that you can use to directly compute numbers of possibilities, because you so often get repeated prime factors. But you can still use prime factors to explain what happens in any particular case, as I've done here. Note that there aren't a lot of ways to _combine_ two or more prime factors to get a single digit: 2 * 2 = 4 2 * 3 = 6 2 * 2 * 2 = 8 3 * 3 = 9 So if you get prime factors of 5 or 7, you can forget about trying to combine them into other digits. And if you ever have a value of p that has a prime factor of more than two digits (for example, p = 26 = 2 * 13), then the answer is easy: there is _no_ way to get the product of the digits to be equal to p. (If you understand why that's the case, then you're well on your way to mastering the idea of prime factoring.) Also, consider what happens if you're looking for 5-digit numbers whose products are 12. You only have three prime factors to work with, so you'll have to use 1 for the remaining digits. I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/