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INTEGER PROBLEM

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by sudhir3127 » Sat Jul 12, 2008 7:28 am
Nice question... My guess is A. 24 can u tell me whats the OA.
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Re: INTEGER PROBLEM

by parallel_chase » Sat Jul 12, 2008 2:10 pm
raj_84 wrote:IF N IS AN INTEGER AND NOT DIVISIBLE BY 3 OR 4
THEN WHAT MUST (N+SIX))*(N+EIGHT)*(N+TEN) DIVISIBLE BY
1.24
2.32
3.96
My take on this None.

If consider N=2, then it is divisible by all of them,

but If we consider N=5 or -5 it is not divisible by either of them.

Simple reasoning is N could be odd or even. If its even the result could be a multiple of any of the three but if its odd it cannot be a multiple of any of the three.

Whats the OA?
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by rs2010 » Sat Jul 12, 2008 8:28 pm
Since 24=2^3*3,32=2^5,96=2^5*3.

So we will always receive one factor of 3 irrespective of value of N even or odd.

We will not get any factor of 2 when N is odd.

When N is even we will at least receive 3 factors of 2 along with one factor of 3.

So, 24 would be obvious choice.
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by parallel_chase » Mon Jul 14, 2008 7:36 am
hemantsood wrote:Since 24=2^3*3,32=2^5,96=2^5*3.

So we will always receive one factor of 3 irrespective of value of N even or odd.

We will not get any factor of 2 when N is odd.

When N is even we will at least receive 3 factors of 2 along with one factor of 3.

So, 24 would be obvious choice.
I understood the part about the even, but what about when N is odd.

The questions asks MUST BE DIVISIBLE irrespective of odd or even.

It would be nice if you could elucidate on this.

Thanks
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