Data Sufficiency

BTGmoderatorLU wrote:Source: GMAT Prep

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 \(3\).

The OA is C

Let's take each statement one by one.

1) \(n\) is not divisible by \(2\).

Case 1: Say n = 1

\((n-1)(n+1)\) = 0. We see that 0 divided by 24 leaves a remainder 0. Thus, r = 0.

Case 1: Say n = 3

\((n-1)(n+1)\) = 2*4 = 8. We see that 8 divided by 24 leaves a remainder 8. Thus, r = 8.

No unique answer. Insufficient.

2) \(n\) is not divisible by \(3\).

Case 1: Say n = 1

\((n-1)(n+1)\) = 0. We see that 0 divided by 24 leaves a remainder 0. Thus, r = 0.

Case 1: Say n = 2

\((n-1)(n+1)\) = 1*3 = 3. We see that 3 divided by 24 leaves a remainder 3. Thus, r = 3.

No unique answer. Insufficient.

(1) and (2) together

From (1) and (2), we have n = 1, 5, 7, 11, and other prime numbers

Case 1: Say n = 1

\((n-1)(n+1)\) = 0. We see that 0 divided by 24 leaves a remainder 0. Thus, r = 0.

Case 2: Say n = 5

\((n-1)(n+1)\) = 4*6. We see that 24 divided by 24 leaves a remainder 0. Thus, r = 0.

Case 3: Say n = 7

\((n-1)(n+1)\) = 6*8. We see that 48 divided by 24 leaves a remainder 0. Thus, r = 0.

Case 4: Say n = 11

\((n-1)(n+1)\) = 10*12. We see that 120 divided by 24 leaves a remainder 0. Thus, r = 0.

So, in each case, we get the remainder 0. Sufficient.

The correct answer: C

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: Manhattan GMAT Classes | GMAT Prep Courses Dubai | LSAT Prep Courses Hong Kong | Singapore SAT Prep Classes | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

n-1, n and n+1 are three consecutive integers. If, as Statement 1 tells us, n is odd, then n-1 and n+1 are both even -- in fact, they are consecutive even numbers. If you take any two consecutive even integers, one of them will always be a multiple of 4, so Statement 1 guarantees that (n-1)(n+1) is divisible by 2*4 = 8. That's not sufficient alone; perhaps one of them is divisible by 3, and (n-1)(n+1) is divisible by 8*3 = 24, and the remainder will be 0 when we divide by 24, or perhaps neither factor is divisible by 3, and we have a nonzero remainder.

From Statement 2, if n is not divisible by 3, then one of n-1 or n+1 must be, since you always find exactly one multiple of 3 among any three consecutive integers. Statement 2 is not sufficient alone, just as Statement 1 was not, but if we use both statements, we know from Statement 1 that (n-1)(n+1) is divisible by 8, and from Statement 2 that it is divisible by 3, so using both Statements (n-1)(n+1) is divisible by 3*8 = 24, and the remainder must be zero when we divide it by 24. So the answer is C.