When A Square Of An Odd Number Is Divided By 8?

1.	When A Square Of An Odd Number Is Divided By 8?
Answer» Let's call the odd number 2a+1, where a is an integer. Then (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1. Since a is odd, a+1 is even. So, 4a(a+1) has to be a multiple of 8 THEREFORE 4a(a+1) + 1 must LEAVE a remainder of 1 when divided by 8. Let's call the odd number 2a+1, where a is an integer. Then (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1. Since a is odd, a+1 is even. So, 4a(a+1) has to be a multiple of 8 Therefore 4a(a+1) + 1 must leave a remainder of 1 when divided by 8.

Answer»

Let's call the odd number 2a+1, where a is an integer.

Then (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1.

Since a is odd, a+1 is even.

So, 4a(a+1) has to be a multiple of 8

THEREFORE 4a(a+1) + 1 must LEAVE a remainder of 1 when divided by 8.

Let's call the odd number 2a+1, where a is an integer.

Then (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1.

Since a is odd, a+1 is even.

So, 4a(a+1) has to be a multiple of 8

Therefore 4a(a+1) + 1 must leave a remainder of 1 when divided by 8.

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