Answer to Question #33200 in Algebra for Bia

Question

The sum of the squares of two digits of a positive integral number is 65 and the number is 9 times the sum of its digits. Find the number.

Answer 1

Let the digits of the number are x and y. Hence we can construct the following equation:
x^2+y^2=65
As x and y are digits, the only option we have is 8 and 1.
So we have two variants for our number: 81 and 18.
Now it's time to use teh second condition.
1) (1+8)*9=18 that's not correct
2) (8+1)*9=81 that's right
So teh required number is 81.