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good one

by vaivish » Tue Sep 02, 2008 9:27 am
. What is the remainder when (n-1)*(n+1) is divided by 24?
(1) When n is divided by 3, the remainder is 1.
(2) n is odd.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Oa is c.
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Re: good one

by sudhir3127 » Tue Sep 02, 2008 9:34 am
vaivish wrote:. What is the remainder when (n-1)*(n+1) is divided by 24?
(1) When n is divided by 3, the remainder is 1.
(2) n is odd.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Oa is c.
Statement 1 . doesnt tell u anything .... insufficent

statement .. odd.. no help... in sufficient

both together ..

n= 7, 13...

(7-1)(7+1)/24 remainder is zero.

(13-1)(13+1)/24 remainder zero.

thus C .

Hope this helps..
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Re: good one

by Stuart@KaplanGMAT » Tue Sep 02, 2008 9:43 am
vaivish wrote:. What is the remainder when (n-1)*(n+1) is divided by 24?
(1) When n is divided by 3, the remainder is 1.
(2) n is odd.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Let's start by rewriting the question!

What is the remainder when (n^2 - 1) is divided by 24?

(1) n/3 has a remainder of 1.

If n = 4, then n^2 - 1 = 15. 15/24 = 0rem15
If n = 7, then n^2 - 1 = 48. 48/24 = 2rem0

insufficient!

(2) n is odd

if n = 3, then n^2 - 1 = 8. 8/24 = 0rem8
we already know that n=7 gives us 2rem0

insufficient!

Together:

n=1, 7, 13, ...

so, n^2 - 1 = 0, 48, 168, ..., all of which are multiples of 24. Sufficient, choose (c).

If you want the "elegant" solution, you need to have a solid understanding of factors (which is why for many test takers, picking numbers is the best way to approach number property yes/no questions).

Back to the original question:

What is the remainder when (n-1)*(n+1) is divided by 24?

24 = 2*2*2*3

So, does (n-1)(n+1) have 2, 2, 2, and 3 among it's prime factors?

(1) n/3 has a remainder of 1.

So, (n-1) is a multiple of 3. (n+1)/3 has a remainder of 2.

All we know for sure is that the expression has a factor of 3, we don't know about the 2s: insufficient.

(2) n is odd.

If n is odd, then n-1 and n+1 are consecutive even integers. All pairs of consecutive even integers have at least 3 factors of 2 between them. (Remember, "0" has an infinite number of factors.)

So, we know that we have our 3 "2"s, but we don't know if we have any "3"s: insufficient.

Together, we know that we have at least 1 factor of 3 and at least 3 factors of 2: sufficient, choose (c).
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by vaivish » Tue Sep 02, 2008 11:12 am
that was awesome reply ...thanks a ton....i will never get wrong on this kind of questions now....hope to c u r replies for future queries ...the d day for me is Oct 5th.....
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by canuckclint » Wed Sep 03, 2008 1:21 pm
It'd be nice to know some simple modular arithmetic. Very useful for remainder questions.


n = 1 (mod 3) (just says n/3 gives 1 as remainder in simple terms)

n^2 -1 = ? (mod 3) need to find ?

n^2 = 1 (mod 3). Square of a number will give us the same remainder.

so n^2 -1 = 0 (mod 3). Just substitue n^2 with 1 from above.

So it n^2-1 is divisible by 3, but what about 24

24 = 3*2^3 (fundamental theorem of arithmetic, just prime factors in gmat)

n^2 - 1 = ? (mod 24). For 2 to be a factor. n^2 -1 must be even.
so n^2 must be odd, so n must be odd.

https://en.wikipedia.org/wiki/Modular_arithmetic
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by Ian Stewart » Wed Sep 03, 2008 4:32 pm
canuckclint wrote:It'd be nice to know some simple modular arithmetic. Very useful for remainder questions.


n = 1 (mod 3) (just says n/3 gives 1 as remainder in simple terms)

n^2 -1 = ? (mod 3) need to find ?

n^2 = 1 (mod 3). Square of a number will give us the same remainder.

so n^2 -1 = 0 (mod 3). Just substitue n^2 with 1 from above.

So it n^2-1 is divisible by 3, but what about 24

24 = 3*2^3 (fundamental theorem of arithmetic, just prime factors in gmat)

n^2 - 1 = ? (mod 24). For 2 to be a factor. n^2 -1 must be even.
so n^2 must be odd, so n must be odd.

https://en.wikipedia.org/wiki/Modular_arithmetic
You've demonstrated that (n-1)(n+1) is divisible by 3, but I don't quite follow the logic in the second part. It appears to me that you are trying to prove that n is odd by assuming the remainder is 0 when (n+1)(n-1) is divided by 24; that is, you are trying to prove Statement 2 is true, if I follow what you're saying. Statement 2 is a fact, and we want to know whether (n+1)(n-1) must be divisible by 8 = 2^3 (since we already know it's divisible by 3). You can certainly do this with modular arithmetic:

If n is odd, either:

n-1 = 0 mod(4) and n+1 = 2 mod(4); that is, n+1 = 0 mod(2)

or

n+1 = 0 mod(4) and n-1 = 2 mod(4); that is, n-1 = 0 mod(2)

So (n+1)(n-1) = 0 mod(8).

Still, one can do the whole problem quite quickly without any modular arithmetic (though modular arithmetic is certainly a useful tool on some questions):

n-1, n, n+1 are three consecutive integers. One of them must be divisible by 3; if n is not, then either n-1 or n+1 is (in fact, n-1 is, from the info given). If n is odd, then n-1 and n+1 are consecutive even integers, so one of them must be divisible by 4, the other by 2. Thus (n-1)(n+1) must be divisible by 3*4*2 = 24.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by dnairo1981 » Thu Oct 02, 2008 5:21 pm
I must be missing something here but can anyone explain why n can not be 1? Would that not mean that the three consecutive integers can be 0, 1, & 2?
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by Stuart@KaplanGMAT » Thu Oct 02, 2008 10:47 pm
dnairo1981 wrote:I must be missing something here but can anyone explain why n can not be 1? Would that not mean that the three consecutive integers can be 0, 1, & 2?
n could indeed be 1 - but I'm not sure why you think that changes anything. In fact, my post stated:
n=1, 7, 13, ...
If n=1, then n^2 - 1 = 0. What's the remainder when 0 is divided by 24? 0.

So, n=1 gives us the same answer to the question as all other possible values of n.
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