You might notice that this is a difference of squares:zagcollins wrote:If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?
1)n is odd
2)n is not divisible by 8
When you divide any integer by 8, the only possible remainders are:sanjaylakhani wrote:but what if N=1, which is possible since it is a positive odd integer, then remainder is 8 -right
Thanks for the great explanation. Can you please explain this a little further. I am trying to understand the concept rather than just the answer to this question.Ian Stewart wrote:If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8.
Yes, this is helpful...thank you.cramya wrote:Anniev,
Take any 2 consecutive even numbers & u will notice:
1) One of the numbers will always be divisible by 4
2)The product will always be divisible by 8 since one number is alreayd divisible by 4 and the other being even will always provide 2 as a factor
Hope this helps!
Regards,
CR
This is what I call mathemajics!Ian Stewart wrote:You might notice that this is a difference of squares:zagcollins wrote:If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?
1)n is odd
2)n is not divisible by 8
n^2 - 1 = (n+1)(n-1)
If n is odd, then n-1 and n+1 are consecutive even integers. If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8. Thus, if we know n is odd, we can be certain that n^2 -1 will be divisible by 8, and r will be zero. 1) is sufficient. 2) is not; n might be even, or might be odd. A.
Only statement 1 is sufficient:zagcollins wrote:If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?
1)n is odd
2)n is not divisible by 8
...I am embarrased....
