[Guest]

Determine all possible value(s) of a positive integer x, such that the sum of the squares of the digits of x is equal to x-2006.

[Guest]

I think these are the only values

2023 & 2083

[superprismatic]

Suppose x has n digits. Then, the sum of the

squares of its digits is less than or equal to

81n (this is attained when all of the n digits

are 9). So, x-2006 is less than or equal to 81n,

i.e., x is less than or equal to 2006+81n. Note

that when n>4, 2006+81n has fewer than n digits

(e.g., if n were 100, 2006+81n=10106 which is

nowhere near being a 100-digit number). Also,

if n were less than 4, x-2006 would be negative

and thereby could not be a sum of squares. We

are forced to conclude that n=4. So, since

0 < x-2006 <= 81*4 from which we get 2006 < x <= 2330.

Trying all 324 of the possible values of x,

we only get the two values 2023 and 2083.

[Guest]

well of course I didn't try the 324 values

I got it by knowing gradually the values of the digits

I let x be a+10b+100c+1000d so a+10b+100c+1000d=2006+a^2+b^2+c^2+d^2

saying that a^2+b^2+c^2+d^2 can be at most if they are 9 to be 324

so x must be more than 2006 by few hundereds so d is now known to be 2 so the equation becomes a+10b+100c=10+a^2+b^2+c^2

then a^2+b^2+c^2 can be at most 243 of course then c must be 2 or less

but when considering that a^2+b^2 is at most 162 so c is either 1 or 0

here the trying goes on

if we considered c=1 we can't solve this equation in natural no.s" a+10b+89=a^2+b^2"

but when considering c=0 the equation becomes a+10b=10+a^2+b^2

trying no.s for b 0,1,3,4,5,6,7,9 a failed to be natural no. and worked only for 2 and 3 to let a=3

that's bring us to 2023 & 2083