(1) n mod 2 != 0
n = 3, n^2-1 = 8. Remainder = 8
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.
(2) n mod 3 != 0
n = 2, n^2-1 = 3. Remainder = 3
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.
Now, putting both together:
n mod 2 != 0
n mod 3 != 0
n^2-1 = (n+1)(n-1)
When n mod 2 != 0, n is odd. So, both n+1 and n-1 are divisible by 2. Hence, (n+1)(n-1) is divisible by 4.
When n mod 3 != 0, either n-1 or n+1 is divisible by 3. (n-1, n and n+1 are 3 consecutive integers).
Also, since n is odd AND not divisible by 3, either n-1 or n+1 is divisible by 4. Now, we accounted for either n-1 or n+1 divisible by 2 using the statement (1). So, (n-1)(n+1) is divisible by 8.
Combining, (n-1)(n+1) is divisible by 8*3 = 24.
Hence, remainder when n^2-1 is divided by 24 is 0.
r=0.
Answer is (C).
Got one from GMATPrep
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
givemeanid
- Master | Next Rank: 500 Posts
- Posts: 277
- Joined: Sun Jun 17, 2007 2:51 pm
- Location: New York, NY
- Thanked: 6 times
- Followed by:1 members
Nice problem... awesome solution... great thinkinggivemeanid wrote:(1) n mod 2 != 0
n = 3, n^2-1 = 8. Remainder = 8
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.
(2) n mod 3 != 0
n = 2, n^2-1 = 3. Remainder = 3
n = 5, n^2-1 = 24. Remainder = 0
Not sufficient.
Now, putting both together:
n mod 2 != 0
n mod 3 != 0
n^2-1 = (n+1)(n-1)
When n mod 2 != 0, n is odd. So, both n+1 and n-1 are divisible by 2. Hence, (n+1)(n-1) is divisible by 4.
When n mod 3 != 0, either n-1 or n+1 is divisible by 3. (n-1, n and n+1 are 3 consecutive integers).
Also, since n is odd AND not divisible by 3, either n-1 or n+1 is divisible by 4. Now, we accounted for either n-1 or n+1 divisible by 2 using the statement (1). So, (n-1)(n+1) is divisible by 8.
Combining, (n-1)(n+1) is divisible by 8*3 = 24.
Hence, remainder when n^2-1 is divided by 24 is 0.
r=0.
Answer is (C).
