# Thread: Sum of squares of digits dividing the number with those digits

1. ## Sum of squares of digits dividing the number with those digits

How many numbers between 1 and 100 (excluded) satisfy that the sum of the squares of its digits divides the number?

Thank you.

2. Originally Posted by guidol92
How many numbers between 1 and 100 (excluded) satisfy that the sum of the squares of its digits divides the number?

Thank you.
Well clearly no one digit number greater than one satisfies this, so let's only look at two digit numbers.

Let $n=10x+y$ where $0\leq x,y \leq 9$

We need to solve the Diophantine Equation $x^2+y^2 = 10x+y$. Before I dive right in, let's see if you can do this.

3. Originally Posted by guidol92
How many numbers between 1 and 100 (excluded) satisfy that the sum of the squares of its digits divides the number?

Thank you.
Well clearly no one digit number greater than one satisfies this, so let's only look at two digit numbers.

Let $n=10x+y$ where $0\leq x,y \leq 9$

We need to solve the Diophantine Equation $\displaystyle x^2+y^2 = 10x+y$. Before I dive right in, let's see if you can do this.

4. Originally Posted by chiph588@
Well clearly no one digit number greater than one satisfies this, so let's only look at two digit numbers.

Let $n=10x+y$ where $0\leq x,y \leq 9$

We need to solve the Diophantine Equation $\displaystyle x^2+y^2 = 10x+y$. Before I dive right in, let's see if you can do this.
Do you notice the word "divides", not "equal to" in x^2+y^2 = 10x+y.

5. Originally Posted by Shanks
Do you notice the word "divides", not "equal to" in x^2+y^2 = 10x+y.
Pardon?

6. A computer search finds the numbers: 1, 10, 20 and 50.

7. @Chip
$x^2 + y^2 | 10x + y$
so
$(x^2 + y^2)k = 10x + y$

8. Originally Posted by elemental
@Chip
$x^2 + y^2 | 10x + y$
so
$(x^2 + y^2)k = 10x + y$
Ohh, I read the question wrong as Shanks was pointing out.