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Value r

by dreamv » Wed Feb 22, 2012 2:25 pm
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 Anurag@Gurome » Wed Feb 22, 2012 7:08 pm
dreamv 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) n is not divisible by 2.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 9, then (n - 1)(n + 1) = 8 * 10 = 80 and r = 8
No definite answer; NOT sufficient.

(2) n is not divisible by 3.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 8, then (n - 1)(n + 1) = 7 * 9 = 63 and r = 15
No definite answer; NOT sufficient.

Combining (1) and (2), we know that n is an odd integer so that it is not divisible by 3.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 5, then (n - 1)(n + 1) = 4 * 6 = 24 and r = 0
If n = 7, then (n - 1)(n + 1) = 6 * 8 = 48 and r = 0
It can be seen that the remainder is always 0; SUFFICIENT.

The correct answer is C.
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by pemdas » Wed Feb 22, 2012 9:52 pm
cannot agree more about statements 1 and 2 taken alone.
st(1) n is odd and (n-1)(n+1) is even, returning numerous remainders. Insufficient.
st(2) n can be even too, so the test put by different even numbers shows Insufficiency.

combining st(1&2): n is odd and not divisible by 3.

the product of two consecutive numbers is not always divisible by 3 and we have little chance to determine divisibility by 24 which contains 3 and 8 or (2^3). The product of three consecutive numbers such as (n-1)n(n+1) is on the contrary always divisible by 3. BUT there's one exception to the product of consecutive numbers divisible by 3, which is based on the value of 'n'. If 'n' is even its function (n-1)(n+1) is always divisible by 3. Yet here, we have 'n' constrained to odd value. Again, there's one more room, if 'n' is odd not divisible by 3 it behaves as 'n' even and is always divisible by 3 with the function (n-1)(n+1).

To list we have the first odd value not divisible by 3 given as 1. (1-1)(1+1) is 0 and is divisible by 3 and 24 as well with remainder=0. The next odd not divisible by 3 will be 5 which is just 24 embedded in function (n-1)(n+1). The next number will be n=7, which within (n-1)(n+1)=6*8 contains at least one 3 (divisibility by 3 is ascertained now) and one even factor with at least three 2's. Exactly what we need one 3 and three 2's to have divisibility by 24. Continuing further we get always at least one 3 and three 2's and have divisibility by 24 for all n=odd values not divisible by 3 through the positive infinity.

This is a full-proof of statements(1&2) combined Sufficiency to answer this DS besides episodic tests by Anurag using numbers 1,5,7.

c
Anurag@Gurome wrote:
dreamv 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) n is not divisible by 2.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 9, then (n - 1)(n + 1) = 8 * 10 = 80 and r = 8
No definite answer; NOT sufficient.

(2) n is not divisible by 3.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 8, then (n - 1)(n + 1) = 7 * 9 = 63 and r = 15
No definite answer; NOT sufficient.

Combining (1) and (2), we know that n is an odd integer so that it is not divisible by 3.
If n = 1, then (n - 1)(n + 1) = 0 and r = 0
If n = 5, then (n - 1)(n + 1) = 4 * 6 = 24 and r = 0
If n = 7, then (n - 1)(n + 1) = 6 * 8 = 48 and r = 0
It can be seen that the remainder is always 0; SUFFICIENT.

The correct answer is C.
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