Remainder Prob
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The only exception to this is 1. But that won't trouble the answer in any way, because in that case, the product will be 0 - (1-1)*(1+1) = 0 thus choice A will hold true.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.
1. If n is odd integer, n = 2k + 1
We have : n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k = 4k(k + 1) is multiple of 8. So r = 0 : Sufficient ---> AD
2. If n = 1 : r = 0
If n = 2 : r = 3
Hence, (2) is insufficient ---> A is correct anwser.
We have : n^2 - 1 = (2k + 1)^2 - 1 = 4k^2 + 4k = 4k(k + 1) is multiple of 8. So r = 0 : Sufficient ---> AD
2. If n = 1 : r = 0
If n = 2 : r = 3
Hence, (2) is insufficient ---> A is correct anwser.
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For the expression (n^2)-1/8
If n is odd, remainder will always be 0
for e.g. 3^2 -1=9-1=8/8. Rem =0
5^2-1=24/8=3. Rem =0
13^2=169-1=168/8/. Rem=0
Thus option 1 is sufficient to answer the question.
Coming to Option 2,
2. n is not divisible by 8
This is not sufficient as n can be any number.
for e.g (9^2) -1=80/8. Rem =0
(10^2) -1=99/8. Rem =3
Thus option 2 cannot provide us the correct answer for the question.
Hence, option 1 is the answer.
If n is odd, remainder will always be 0
for e.g. 3^2 -1=9-1=8/8. Rem =0
5^2-1=24/8=3. Rem =0
13^2=169-1=168/8/. Rem=0
Thus option 1 is sufficient to answer the question.
Coming to Option 2,
2. n is not divisible by 8
This is not sufficient as n can be any number.
for e.g (9^2) -1=80/8. Rem =0
(10^2) -1=99/8. Rem =3
Thus option 2 cannot provide us the correct answer for the question.
Hence, option 1 is the answer.
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If n is odd it can be expressed as 2k+1 for any integer k.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
So, n^2-1 = (2k+1)^2-1 =4k^2+4k=4k(k+1)=8m (where m is another integer)=> r=0 for all odd n.
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a. if n is odd then n must be 1,3, 5,7,9 then the last digit of n^2 and (n^2-1) must be 1,9,5,9 and 0,8,4,8.
Let (n^2-1) / 8 means that (n^2-1): 2:2:2.
we have the number with last digit 8 is always divisible by 2 ( for last digit only 8:2=4:2=2:2=1) ; the number with last digit 0 will have remainder 1 after dividing by 2:2:2, the number with last digit 4 will have remainder 1 after dividing by 2:2:2.
then the remander will be always 1, A is sufficient
b.n is not divisible by 8 then n can be odd or even => B is insufficient.
the answer is A
Let (n^2-1) / 8 means that (n^2-1): 2:2:2.
we have the number with last digit 8 is always divisible by 2 ( for last digit only 8:2=4:2=2:2=1) ; the number with last digit 0 will have remainder 1 after dividing by 2:2:2, the number with last digit 4 will have remainder 1 after dividing by 2:2:2.
then the remander will be always 1, A is sufficient
b.n is not divisible by 8 then n can be odd or even => B is insufficient.
the answer is A