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?Number plugging method:\((n-1)(n+1)=n^2-1\)

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if \(n=1\), then \(n^2-1=0\) and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if \(n=3\), then \(n^2-1=8\) and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if \(n=1\), then \(n^2-1=0\), so remainder is 0 but if \(n=2\), then \(n^2-1=3\) and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3:

\(n=1\) --> \(n^2-1=0\) --> remainder 0;

\(n=5\) --> \(n^2-1=24\) --> remainder 0;

\(n=7\) --> \(n^2-1=48\) --> remainder 0;

\(n=11\) --> \(n^2-1=120\) --> remainder 0.

Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:(1) n is not divisible by 2. Insufficient on its own, but this statement says that \(n=odd\) --> \(n-1\) and \(n+1\) are

consecutive even

integers --> \((n-1)(n+1)\) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From

consecutive even

integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either \(n-1\) or \(n+1\) must be divisible by 3 (as \(n-1\), \(n\), and \(n+1\) are

consecutive integers, so one of them must be divisible by 3, we are told that it's not \(n\), hence either \(n-1\) or \(n+1\)).

(1)+(2) From (1) \((n-1)(n+1)\) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by \(8*3=24\), which means that remainder upon division \((n-1)(n+1)\) by 24 will be 0. Sufficient.

Answer: C.

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