If n is a positive integer and r is the remainder when n^21 is divided by 8, what is the value of r?
1)n is odd
2)n is not divisible by 8
Remainder Prob

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IMHO the answer is A.
1) Just plug in odd numbers,
n=1 : (1^2  1)/8 = 0
n=3 : (3^2  1)/8 = 0
n=5 : (5^2  1)/8 = 0
n=7 : (7^2  1)/8 = 0
after this I assume that r is always 0, so it´s sufficient
2) Again plug some numbers
n=1 : (1^2  1)/8 = 0
n=2 : (2^2  1)/8 = 3
upz, not sufficient.
1) Just plug in odd numbers,
n=1 : (1^2  1)/8 = 0
n=3 : (3^2  1)/8 = 0
n=5 : (5^2  1)/8 = 0
n=7 : (7^2  1)/8 = 0
after this I assume that r is always 0, so it´s sufficient
2) Again plug some numbers
n=1 : (1^2  1)/8 = 0
n=2 : (2^2  1)/8 = 3
upz, not sufficient.
Last edited by augusto on Mon Jul 28, 2008 12:21 pm, edited 1 time in total.
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 Ian Stewart
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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^21 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)(n1)
If n is odd, then n1 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.
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but what if N=1, which is possible since it is a positive odd integer, then remainder is 8 right
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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
0, 1, 2, 3, 4, 5, 6 and 7
The remainder can never be 8 when you divide by 8. In addition, 0 is divisible by every positive integer, so the remainder is 0 when you divide 0 by 8 (or by any other positive integer).
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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.
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
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
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^21 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)(n1)
If n is odd, then n1 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.

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 Abhishek009
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Only statement 1 is sufficient:zagcollins wrote:If n is a positive integer and r is the remainder when n^21 is divided by 8, what is the value of r?
1)n is odd
2)n is not divisible by 8
Checked with the following as n :
n = 3
n = 5
n = 7
However we can't arrive the answer using statement 2 alone.
Abhishek
If n is a positive integer and r is the remainder when (n^2)1 divided by 8 ,what is the value of r?
I do not understand how the answer is a. i did this to get e as my answer:
1) n is not odd
n q r
4 1 7
6 4 3
8 7 7
This statement is insufficient because it doesnt definitively give you one answer for r. or so i thought
2) n is not divisible by 8
n q r
12 17 7
10 12 3
7 6 0
this again doesnt definitively give you one value for r so i said this was also insufficient.
1and 2) n is not odd and it is not divisible by 8
n q r n q r
2 0 3 6 4 3
4 1 7 10 12 3
when i tried twelve it also gave me a remainder of seven so since r can be three or seven this is also not enough. is there something im missing?
I do not understand how the answer is a. i did this to get e as my answer:
1) n is not odd
n q r
4 1 7
6 4 3
8 7 7
This statement is insufficient because it doesnt definitively give you one answer for r. or so i thought
2) n is not divisible by 8
n q r
12 17 7
10 12 3
7 6 0
this again doesnt definitively give you one value for r so i said this was also insufficient.
1and 2) n is not odd and it is not divisible by 8
n q r n q r
2 0 3 6 4 3
4 1 7 10 12 3
when i tried twelve it also gave me a remainder of seven so since r can be three or seven this is also not enough. is there something im missing?