Thread: Integer Problem

Take an odd integer; square it; then divide by 8. Can the remainder by 3? Can it be 7? (of course, just "yes" or "no" is not an acceptable answer. You must explain why the proposed remainder can or cannot occur.)

Please Help...

I have this so far:

Ex. (3)(3) = 9/8 = 1.125
Wanted to try and get to 3 so I went up with my integer

Ex. (7)(7) = 49/8 = 6.125

Ex. (9)(9) = 81 / 8 = 10.125

I am assuming that this will not work....but am not certain. Does anyone have any other thoughts?

Originally Posted by matgrl

Take an odd integer; square it; then divide by 8. Can the remainder by 3? Can it be 7? (of course, just "yes" or "no" is not an acceptable answer. You must explain why the proposed remainder can or cannot occur.)

Let odd integer = 2x + 1

(2x + 1)^2 / 8
= (4x^2 + 4x + 1) / 8
= [4x(x + 1) + 1] / 8
= 4x(x + 1) / 8 + 1/8
= x(x + 1) / 2 + 1/8

Since x(x+1) is always even, being the product of 2 consecutive integers,
then x(x+1) / 2 is always an integer, so 1/8 is always remainder: that's the .125 you were getting.

This is wonderful thank you! Could you show me how you factored 4x^2 + 4x + 1? I am not sure how to do this. Also could you explain to me how you got 1/8 as the remainder? Thank you for your time. I appreciate it

An integer when divided by another always leaves an integer remainder. You are confused between remainders and quotients. I think you should work like this:

Any odd number has one of the following forms :

a) 8n - 3

b) 8n - 1

c) 8n + 1

d) 8n + 3

Squaring these we get the following forms respectively:

a) 8n + 9

b) 8n + 1

c) 8n + 1

d) 8n + 9

9 = 1 (mod 8)

So an odd square can have only remainder 1 when divided by 8.