If x is a positive integer, what is the remainder when (x+1)(x-1) is divided by 24?
I) x is odd
2) x is not divisible by 3
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- JadranLee
- ManhattanGMAT Staff
- Posts: 108
- Joined: Mon Aug 07, 2006 10:33 am
- Location: Chicago, IL
- GMAT 5/18
Van D,
My method is certainly not the most efficient, but I believe I derived the right answer and it too me 2 minutes exactly (so not too bad).
First, I started with Statement I:
I tried x = 3, x = 5, x = 7, and x = 9. When x = 3, the remainder is 8; x = 5, remainder is 0; x = 7, remainder is 0; x = 9, remainder = 8. Therefore, insufficient.
Then, Statement II:
I tried x = 4, which gave a remainder of 20, and x = 8, which gave a remainder of 6. Therefore, insufficient.
Together, x has to be odd and not divisible by 3. As i already know x = 5 and x = 7 gives a remainder of 0, I tried a couple of other odd numbers not divisible by 3 (1 and 11). Both of these also produced remainders of 0, so I concluded that C was the correct answer.
I think there is a better/faster way to do this, using prime boxes, but I am not too confident using that method. Hope this helps!
My method is certainly not the most efficient, but I believe I derived the right answer and it too me 2 minutes exactly (so not too bad).
First, I started with Statement I:
I tried x = 3, x = 5, x = 7, and x = 9. When x = 3, the remainder is 8; x = 5, remainder is 0; x = 7, remainder is 0; x = 9, remainder = 8. Therefore, insufficient.
Then, Statement II:
I tried x = 4, which gave a remainder of 20, and x = 8, which gave a remainder of 6. Therefore, insufficient.
Together, x has to be odd and not divisible by 3. As i already know x = 5 and x = 7 gives a remainder of 0, I tried a couple of other odd numbers not divisible by 3 (1 and 11). Both of these also produced remainders of 0, so I concluded that C was the correct answer.
I think there is a better/faster way to do this, using prime boxes, but I am not too confident using that method. Hope this helps!
- dbernst
- ManhattanGMAT Staff
- Posts: 300
- Joined: Mon May 09, 2005 9:03 am
Van D,
Though you could consider dealing with prime factors, I personally would have chosen the plugging in method as well. First, I would rephrase the questions to What is the remainder when x^2 - 1 is divided by 24.
For statement (1) try the first few odd integers:
-When x = 1, r=0
When x=3, r=8
Insufficient
For statement (2), try any integer not divisible by three
-when x = 1, r=0
-when x = 2, r=3
Insufficient
Together, you already have
-when x = 1, r=0
Try
-when x = 5, r=0
-when x = 7, r = 0
-when x = 11, r = 0
-when x = 13, r = 0
Together, the first five options for x all provide a remainder of zero. Choose C.
With prime factors, the factors of 24 are 2*2*2*3
When x is odd and NOT divisible by 3, both of the adjacent numbers (x-1) and (x+1) will be even and one will be divisible by 4. Thus, your three powers of 2 are present. Furthermore, one of the two adjacent numbers must be divisible by 3, since you have three consecutive integers (one of every three consecutive integers is divisible by 3) and x is explicitly NOT the one divisible by 3. Therefore your power of three is covered. Since (x-1)(x+1) must include 2*2*2*3, the product must be divided evenly by 24, leaving a remainder of zero.
The correct answer is C
Though you could consider dealing with prime factors, I personally would have chosen the plugging in method as well. First, I would rephrase the questions to What is the remainder when x^2 - 1 is divided by 24.
For statement (1) try the first few odd integers:
-When x = 1, r=0
When x=3, r=8
Insufficient
For statement (2), try any integer not divisible by three
-when x = 1, r=0
-when x = 2, r=3
Insufficient
Together, you already have
-when x = 1, r=0
Try
-when x = 5, r=0
-when x = 7, r = 0
-when x = 11, r = 0
-when x = 13, r = 0
Together, the first five options for x all provide a remainder of zero. Choose C.
With prime factors, the factors of 24 are 2*2*2*3
When x is odd and NOT divisible by 3, both of the adjacent numbers (x-1) and (x+1) will be even and one will be divisible by 4. Thus, your three powers of 2 are present. Furthermore, one of the two adjacent numbers must be divisible by 3, since you have three consecutive integers (one of every three consecutive integers is divisible by 3) and x is explicitly NOT the one divisible by 3. Therefore your power of three is covered. Since (x-1)(x+1) must include 2*2*2*3, the product must be divided evenly by 24, leaving a remainder of zero.
The correct answer is C
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