Sum of IntegersDate: 07/03/2001 at 03:57:14 From: Mark Subject: Sum of integers How many integers are 13 times the sum of their digits? Date: 07/03/2001 at 13:12:33 From: Doctor Rob Subject: Re: Sum of integers Thanks for writing to Ask Dr. Math, Mark. Suppose the number has four or more digits. Then 13*(a+b+c+d+...) = a + 10*b + 100*c + 1000*d + ..., 12*a + 3*b = 87*c + 987*d + ..., 4*a + b = 29*c + 329*d + .... Now all the digits a, b, c, d, ..., are in the set {0,1,2,...,9}, so 45 = 4*9 + 9 >= 4*a + b = 29*c + 329*d + ... >= 0. This implies that d and all higher-order digits must be zero, and c = 0 or 1. If c = 0, then both a and b must also be 0, a contradiction (or a solution, if you allow zero). Thus we have to solve 4*a + b = 29, with integers a and b between 0 and 9 inclusive. This I leave to you. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 07/03/2001 at 13:14:13 From: Doctor Ian Subject: Re: Sum of integers Hi Mark, I'm going to assume that we can ignore negative integers, since the sum of the digits of a negative integer is the same as the sum of the digits of the corresponding positive integer. For a one-digit integer, 'a', the condition is a = 13(a) This is true for a = 0, but not for any other digit. For a two-digit integer 'ab', the condition is 10a + b = 13(a + b) 10a + b = 13a + 13b 0 = 3a + 12b This is only true when a = 0 and b = 0. For a three-digit integer 'abc', the condition is 100a + 10b + c = 13(a + b + c) = 13a + 13b + 13c 0 = -87a + 3b + 12c Again, this is true when a = 0, b = 0, and c = 0, and not for any other set of digits. A reasonable guess would be that 0 is the only number that satisfies the condition, but we haven't proved anything. If you want to try to prove it, proof by induction would probably be the way to proceed. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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