10-Digit Number PuzzleDate: 02/10/2001 at 08:18:42 From: ihsan Subject: Turkish math olympiads - May "abcdefghij" is a ten-digit-number. All of the digits (a,b,c,d...) are different from each other. If 11111 divides it evenly, how many possibilities are there for abcdefghij? Date: 02/12/2001 at 13:19:13 From: Doctor Rob Subject: Re: Turkish math olympiads - May Thanks for writing to Ask Dr. Math, Ihsan. If all digits are different, and there are ten of them, then all the numbers from 1 to 10 must appear. That implies that the digits of abcdefghij add up to 45 = 0 + 1 + 2 + ... + 9. That implies that abcdefghij is divisible by 9. Since 9 and 11111 are relatively prime, their least common multiple is 9*11111 = 99999, and so that must divide evenly into abcdefghij. Now consider what number 99999 must be multiplied by. It must be larger than abcde, because abcde*100000 = abcde00000 < abcdefghij. It can't be abcde+2 (or larger), because: 99999*(abcde+2) = (100000-1)*(abcde+2) = abcde00000 + 199998 - abcde Looking at the 10000's digit of that, 0 + 9 - e would not require a borrow, regardless of the value of e. Therefore the 100000's digit would be e + 1, which doesn't equal the 100000's digit of abcdefghij. That implies that: abcdefghij = 99999*(abcde+1) = (100000-1)*(abcde+1) abcde00000 + fghij = abcde00000 + (99999 - abcde) fghij = 99999 - abcde Then the following equations must all hold: f = 9 - a g = 9 - b h = 9 - c i = 9 - d j = 9 - e Now there are nine choices for a (it cannot be 0), and then f is known. That leaves eight choices for b, and then g is known. That leaves six choices for c, and then h is known. That leaves four choices for d, and then i is known. That leaves two choices for e, and then j is known. Thus the total number of such numbers abcdefghij is 9*8*6*4*2 = 3456. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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