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value of r

by daretodream » Fri Feb 19, 2010 3:47 am
If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

1) n is not divisible by 2.
2) n is not divisible by 3.
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by ajith » Fri Feb 19, 2010 4:42 am
daretodream wrote:If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

1) n is not divisible by 2.
2) n is not divisible by 3.
1) let n = 3 the remainder 8/24 = 8 Let n =5 the remainder = 0 ; Not sufficient
2) n= 2 the remainder = 3 n =4 remainder = 15 ; Not sufficient

Combining n =5, remainder = 1; n =7 remainder =0; n = 11 , remainder = 0; n =13 remainder =168

This is because if n is divisible by 2 n+1 and n-1 both are divisible by 2 and( n+1)* (n-1) is a multiple of 8 and one of them is a multiple of 3 if n is an odd number and not a multiple of 3

Sufficient

C
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by thephoenix » Fri Feb 19, 2010 10:14 am
daretodream wrote:If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

1) n is not divisible by 2.
2) n is not divisible by 3.
IMO it should be C,

1. Is not suff because it only tells us that the N is odd and we can diff possibilities for it for e.g n= 9 which means (n-1)(n+1)=8x10=80 which is gives a remainder of 8 and we can also have N= 5 where (n-1)(n+1)= 4x6=24 remainder =0

2. Similarly this is not suff; and we can have diff values like n=8 , (n-1)(n+1)=7x9 = 63, hence r= 15 and we have n=5 as well where r =0.

But we combine both these 2 we would get values with the pattern as N=5, 7, 11, 13, 17, 19, 23, .... and in all these cases (n-1)(n+1) will give us the remainder as 0 only ((n-1)(n+1) will have multiples of 2, and 3).
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