1. Possible remainders

What are possible remainders when dividing with 9 squares of natural numbers?

2. Originally Posted by OReilly
What are possible remainders when dividing with 9 squares of natural numbers?
You can attack the problem with Division Algorithm.

Meaning any number must be exactly in one of these forms,
$9k$
$9k+1$
$9k+2$
.........................
$9k+8$

When you square then, you get, the only possibilities,
$9k$
$9k+1$
$9k+4$
$9k+7$
Thus, remainders are 0,1,4,7

3. I must say I didn't think of that!
This problem has destroyed me! Whole day I couldn't solve it. Problems with divisibility are giving me most hard times and usually I can't solve them!