[Guest]

(A) Determine all possible value(s) of a positive integer x, such that the sum of squares of the digits in x is equal to 2009-x.

(B) Determine all possible value(s) of a positive integer y, such that the sum of squares of the digits in y is equal to 2005-y.

In each of (A) and (B), prove that there are no other value(s) that exist.

[Guest]

I pretty much brute-forced it, but I eliminated a lot of possibilities, so I didn't have to try very many numbers.

First of all, note that x itself must be less than 2009. First, I tried values of x from 2000 through 2009, and none of them worked. Next, I supposed that we have a 4-digit number, ABCD (A, B, C, and D are the digits, and are thus integers from 0-9 inclusive). We can then say...

1000A + 100B + 10C + D + A² + B² + C² + D² = 2009.

A(1000 + A) + B(100 + B) + C(10 + C) + D(1 + D) = 2009

If A = 0, then we have B(100 + B) + C(10 + C) + D(1 + D) = 2009

Unfortunately, with the constraint that B, C, and D are at most 9, B(100 + B) + C(10 + C) + D(1 + D) is at most 1242, so A must be equal to 1. Knowing that A = 1, we now have:

1(1001) + B(100 + B) + C(10 + C) + D(1 + D) = 2009

B(100 + B) + C(10 + C) + D(1 + D) = 1008

Since C and D are at most 9, we can also say C(10 + C) + D(1 + D) is at most 9*19 + 9*10 = 261, which means that B(100 + B) is at least 1008 - 261 = 747. This means that B is 7, 8, or 9. Let's try all three:

B = 7: C(10 + C) + D(1 + D) = 259. C = D = 9 does not work here, and C = 9 and D = 8 is too small. No suitable solutions here.

B = 8: C(10 + C) + D(1 + D) = 144. Since D(1 + D) must be even, 144 - D(1 + D) = C(10 + C) is also even, so C is even. C = 8 yields D = 0, and upon inspection, no other values of C will work.

B = 9: C(10 + C) + D(1 + D) = 27. C(10 + C) is at most 27, so C can only be 0, 1, or 2. None of these values of C yields a suitable solution for D.

The only suitable quadruplet (A,B,C,D) is (1,9,1,3), so the only value of x that will work is x = 1880.

The math is basically the same for the 2005 problem...in fact, I just copied and pasted the above solution and changed a few numbers.

First of all, note that x itself must be less than 2005. First, I tried values of x from 2000 through 2005, and none of them worked. Next, I supposed that we have a 4-digit number, ABCD (A, B, C, and D are the digits, and are thus integers from 0-9 inclusive). We can then say...

1000A + 100B + 10C + D + A² + B² + C² + D² = 2005.

A(1000 + A) + B(100 + B) + C(10 + C) + D(1 + D) = 2005

If A = 0, then we have B(100 + B) + C(10 + C) + D(1 + D) = 2005

Unfortunately, with the constraint that B, C, and D are at most 9, B(100 + B) + C(10 + C) + D(1 + D) is at most 1242, so A must be equal to 1. Knowing that A = 1, we now have:

1(1001) + B(100 + B) + C(10 + C) + D(1 + D) = 2005

B(100 + B) + C(10 + C) + D(1 + D) = 1004

Since C and D are at most 9, we can also say C(10 + C) + D(1 + D) is at most 9*19 + 9*10 = 261, which means that B(100 + B) is at least 1004 - 261 = 743. This means that B is 7, 8, or 9. Let's try all three:

B = 7: C(10 + C) + D(1 + D) = 255. C = D = 9 does not work here, and C = 9 and D = 8 is too small. No suitable solutions here.

B = 8: C(10 + C) + D(1 + D) = 140. Since D(1 + D) must be even, 144 - D(1 + D) = C(10 + C) is also even, so C is even. Trying all possible even values for C yields no suitable solutions.

B = 9: C(10 + C) + D(1 + D) = 23. C(10 + C) is at most 23, so C can only be 0 or 1. C = 1 yields D = 3. C = 0 yields no suitable solution.

The only suitable quadruplet (A,B,C,D) is (1,9,1,3), so the only value of x that will work is x = 1913.