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## Remainder Prob

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mana6 Just gettin' started!
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Sat Jan 08, 2011 4:26 pm
i see my own mistake i misread statement one its n is odd. if i use this statement now i do get that r should be zero and this statement is then sufficient.

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niazsna786 Just gettin' started!
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Wed Mar 23, 2011 5:29 am
The answer is very clear A
With 2 we can say value of r can be 0 1 2 3 4 5 6 or 7
but with 1 we can say for sure it is 0
i.e.
n = 1 expr (1-1)/8 , r= 0
n = 3 expr (9-1)/8, r= 0
n = 5 expr (25-1)/8, r= 0

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Tue May 17, 2011 7:40 am
I see that in this case, it is quicker to use number properties to get the answer. But, I went around plugging in values, which was obviously more time consuming.
At the time I saw this question, it seemed to me a clear candidate for plugging values!

Is there any means by which I could have identified the mode of attack better?

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coolly01 Just gettin' started!
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Wed May 18, 2011 4:05 am
Ian Stewart wrote:
zagcollins wrote:
If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?

1)n is odd
2)n is not divisible by 8
You might notice that this is a difference of squares:

n^2 - 1 = (n+1)(n-1)

If n is odd, then n-1 and n+1 are consecutive even integers. If you take any two consecutive even integers, one of them will be divisible by 4, the other not, so their product must be divisible by 8. Thus, if we know n is odd, we can be certain that n^2 -1 will be divisible by 8, and r will be zero. 1) is sufficient. 2) is not; n might be even, or might be odd. A.
Hi Steward,
I have one stupid question about divisibility and remainder and it really confuses me.
About divisibility: It's clear that 10 is indivisible by 3 and the remainder is 1 (because 1/3 is infinitive)
However: Is 12 divisible by 8 as the result is 1.5 (4 is divided by 8)?
What does divisibility mean? (Does it refer to integer or infinity?)
Kelly

sayanchakravarty Just gettin' started!
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Mon May 30, 2011 10:23 pm
A

mandeepak Rising GMAT Star
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Mon Jun 06, 2011 8:03 pm
IMO - A

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Mon Jun 06, 2011 9:33 pm
If n is a positive integer and r is the remainder when n^2-1 is divided by 8, what is the value of r?

1)n is odd
2)n is not divisible by 8

(n-1)(n+1) divided by 8
a) n is odd, and thus (n-1),(n+1) are 2 consecutive even numbers. thus one of them is divisible by 4 and one of them is divisible by 2. Thus product is divisible by 8. Sufficient (r=0)
b)n is not divisible by 8. But it can be even and odd (even->2,odd->3)
Insufficient.
IMO A

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Ian Stewart GMAT Instructor
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Tue Jun 07, 2011 12:55 pm
coolly01 wrote:
I have one stupid question about divisibility and remainder and it really confuses me.
About divisibility: It's clear that 10 is indivisible by 3 and the remainder is 1 (because 1/3 is infinitive)
However: Is 12 divisible by 8 as the result is 1.5 (4 is divided by 8)?
What does divisibility mean? (Does it refer to integer or infinity?)
I just saw this now. We say that an integer x is divisible by an integer y if x/y is an integer. So 14 is divisible by 7, because 14/7 = 2, which is an integer. However 12 is *not* divisible by 8, since 12/8 = 1.5 is not an integer.

I think you might be confusing the concept of divisibility with that of terminating/repeating decimals. A number is only divisible by another if you get no decimal part at all when you divide.

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coolly01 Just gettin' started!
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Tue Jun 07, 2011 8:57 pm
Ian Stewart wrote:
coolly01 wrote:
I have one stupid question about divisibility and remainder and it really confuses me.
About divisibility: It's clear that 10 is indivisible by 3 and the remainder is 1 (because 1/3 is infinitive)
However: Is 12 divisible by 8 as the result is 1.5 (4 is divided by 8)?
What does divisibility mean? (Does it refer to integer or infinity?)
I just saw this now. We say that an integer x is divisible by an integer y if x/y is an integer. So 14 is divisible by 7, because 14/7 = 2, which is an integer. However 12 is *not* divisible by 8, since 12/8 = 1.5 is not an integer.

I think you might be confusing the concept of divisibility with that of terminating/repeating decimals. A number is only divisible by another if you get no decimal part at all when you divide.
I'm clear now.
Thanks so much Stewart
Kelly

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Thu Jul 14, 2011 2:58 am
n^2 - 1 can be represented as (n-1)(n+1).
now if n = odd n-1 and n+1 represent two consecutive even number.
so n - 1 = 2 * x where x = any integer
n + 1 = (n-1) + 2 = 2*x + 2 = 2 (x+1)

Case1: if x = odd, x+1 will be even. So n+1 will be divisible by 2 * (Even Number) which means 4
Also n-1 will be divisible by 2 only. since x = odd.
Combining (n-1)(n+1) will be divisible by 4*2 = 8.
Case2: If x = even, x+1 will be odd. So n+1 will be divisible only by 2.
n-1 will be divisible by 2 * Even Number = 4
Combining (n-1)(n+1) will be divisible by 4*2 = 8.

Hence A) is the correct answer.

breakkgmat Just gettin' started!
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Thu Jul 14, 2011 10:40 am
As statement 1 says-- n is odd..so just pick any odd value of n & you will see that the reminder is 0.
for example. n is 5. N^2 1 is 24 which is divisible by 8.n is 7 n^2-1 is 48 which is also divisible by 8 with reminder of 0.so you know r is always 0..\
St2.does not tell us anything about r or more..

OA - A

shingik Just gettin' started!
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Sat Jul 16, 2011 5:38 pm
1) is insufficient because n can be any odd number
2) is insufficient because n can be any number not divisible by 8 but it does rule out 1 as a possiblity. 0 is divisble by 8.
Together they are sufficent because the remainder is always zero for any odd number squared minus 1 divided by 8. (Performed)operation with 4 consecutive odd numbers.
Therefore I chose C and my answer.

shingik Just gettin' started!
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Sat Jul 16, 2011 5:46 pm
shingik wrote:
1) is insufficient because n can be any odd number
2) is insufficient because n can be any number not divisible by 8 but it does rule out 1 as a possiblity. 0 is divisble by 8.
Together they are sufficent because the remainder is always zero for any odd number squared minus 1 divided by 8. (Performed)operation with 4 consecutive odd numbers.
Therefore I chose C and my answer.
Just noticed my own dumb mistake --- A is sufficient.

Deependra1 Rising GMAT Star
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Mon Aug 29, 2011 6:36 am

Should be correct this time around

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