A Number Digits Puzzle
Date: 02/23/2001 at 22:00:44 From: Nikhil Promise Subject: Number Theory Determine all positive integers with the property that they are one more than the sum of the squares of their digits in base 10.
Date: 02/26/2001 at 09:50:12 From: Doctor Rob Subject: Re: Number Theory Thanks for writing, Nikhil. Let F(n) be one more than the sum of the squares of the digits of n. You can prove pretty easily that there are no numbers n with F(n) = n and d >= 4 digits, because d-digit numbers are at least 10^(d-1), but F(n) is at most d*9^2 + 1 = 81*d + 1, and for those d's, 10^(d-1) > 81*d + 1. This would imply that: n >= 10^(d-1) > 81*d + 1 >= F(n) = n so n > n, a contradiction. That just leaves 999 cases to consider. Now you can eliminate all three-digit numbers greater than 244, because: 244 = 81*3 + 1 >= F(n) = n That allows you to eliminate all three-digit numbers greater than 164, because the largest F(n) could be for n <= 244 is: F(199) = 1^2 + 9^2 + 9^2 + 1 = 164 That allows you to eliminate all three-digit numbers greater than 108, because: F(159) = 1^2 + 5^2 + 9^2 + 1 = 108 That allows you to eliminate all three-digit numbers, because: F(108) = 1^2 + 0^2 + 8^2 = 65 Obviously there are no one-digit numbers with F(n) = n, since the equation a^2 + 1 = a has no real roots. That means that you are trying to solve a^2 + b^2 + 1 = 10*a + b Multiply by 4, bring all terms over to the left-hand side, and complete the squares on a and b: 4*a^2 - 40*a + 4*b^2 - 4*b + 4 = 0 4*a^2 - 40*a + 100 + 4*b^2 - 4*b + 1 = 97 (2*a-10)^2 + (2*b-1)^2 = 97 Now there is just one way of writing the prime 97 as the sum of two squares. The even square must be (2*a-10)^2 and the odd square must be (2*b-1)^2, and that will tell you the possible values of a and b. It looks as if there are just two solutions. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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