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statement (1) says 2 is not a factor of n. let n = 3, (n-1)(n+1) = 2*4/24 = 8/24 (remainder 8). let n be 5. then (n-1)n+1)/24 = 4* 6/24 (remainder zero). different remainders, insuffiecient.

statement 2 also insuffiecient, plug in value of n as 2 and 5 and we get different remainders.

combining both statements we have to take n as integer which is not a factor of 2 and 3. let us take n= 5. then (5-1)(5 +1)/24 remainder zero. again take n 7, then (7-1)(7+1)/24 = 8*6/24. remainder zero. so combining statement1 and 2 is sufficient to let us know that remainder of (n-1)(n+1)/24 is always going to be zero when n is not a factor of 2 or 3.

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) 2 is not a factor of n (2) 3 is not a factor of n

Another approach:

(n-1), n and (n+1) are consecutiveintegers.

If 2 is not a factor of n (i.e. n is odd), it must be a factor of (n-1) and (n+1) (the numbers around n must be even). Also, out of any two consecutive even numbers, one has to be divisible by 4 because every alternate multiple of 2 is divisible by 4. Hence, (n-1)*(n+1) must be divisible by 8.

Out of any 3 consecutiveintegers, one has to be divisible by 3 because every third integer is a multiple of 3. Out of (n-1), n and (n+1), one has to be divisible by 3. If n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3. Hence, (n-1)*(n+1) must be divisible by 3.

Using both statements, (n-1)*(n+1) must be divisible by 8*3 = 24. Remainder must be 0.

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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 consecutiveintegers, 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.

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