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# GMATPREP EXAM 1 (2)

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GMATPREP EXAM 1 (2) [#permalink]  18 Sep 2007, 22:23
Can someone explain the logic in this question? How do you approach this divisibility question and all other for that matter?

thanks
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This one took me quite a bit of time =( bout 3 1/2 min. thats no good.

Owell.

rewrite the statment like this n^2-1/24.

Then _____
24|n^2-1 (if this doesn't show the way I want it to, essentially, use the long division box)

S1: since n cant be divisible by 2 try values for n that arent divisible by 2.

5^2 =25. ---> 25-1/24 R here will be 0

Try now finding a value you know won't be divisible by 24.

9^2 81-1/24 this is def. not going to have a R of 0. So insuff.

S2:
cannot be divisible by 3. So again try 5^2-1/24 R=0.

Try finding a value that u know won't leave a R of 0.

10^2-1/24 99/24 Def R does not equal 0.

S1&S2:
Realize here that the following values for n are not possible: 2,3,4,6,8,9,10 and anything divisible by 2 or 3! So this essentially leaves primes of 5,7,11,13 etc...

Try a couple and determine ans from here as I was already at 3min!

Try a few values to see if this works again 5^2-1/24 24/24 R=0

7^2-1/24 R=0.

11^2-1/24 R=0.

Ans C.
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Re: GMATPREP EXAM 1 (2) [#permalink]  18 Sep 2007, 23:56
ericuva wrote:
Can someone explain the logic in this question? How do you approach this divisibility question and all other for that matter?

thanks

I plugged in numbers to solve this one.

Stat 1: if n=3 remainder = 0, if n=9 remainder=1, Insuff.

Stat 2: if n=5, remainder=0, if n=10, remainder=3, Insuff.

Stat 1 & 2: n=5, 7, 11, 13, remainder = 0.

Re: GMATPREP EXAM 1 (2)   [#permalink] 18 Sep 2007, 23:56
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