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

• Nov 22nd 2010, 06:26 PM
guidol92
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.
• Nov 22nd 2010, 10:03 PM
chiph588@
Quote:

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

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.
• Nov 22nd 2010, 10:03 PM
chiph588@
Quote:

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

We need to solve the Diophantine Equation $\displaystyle \displaystyle x^2+y^2 = 10x+y$. Before I dive right in, let's see if you can do this.
• Nov 22nd 2010, 11:14 PM
Shanks
Quote:

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

We need to solve the Diophantine Equation $\displaystyle \displaystyle x^2+y^2 = 10x+y$. 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.
• Nov 23rd 2010, 03:40 PM
chiph588@
Quote:

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

Pardon? (Thinking)
• Nov 27th 2010, 02:38 PM
qmech
A computer search finds the numbers: 1, 10, 20 and 50.
• Nov 28th 2010, 06:05 AM
elemental
@Chip
$\displaystyle x^2 + y^2 | 10x + y$
so
$\displaystyle (x^2 + y^2)k = 10x + y$
• Nov 28th 2010, 09:22 PM
chiph588@
Quote:

Originally Posted by elemental
@Chip
$\displaystyle x^2 + y^2 | 10x + y$
so
$\displaystyle (x^2 + y^2)k = 10x + y$

Ohh, I read the question wrong as Shanks was pointing out.