If n is odd, then n(n*n - 1) is divisible by: 1. 9 2. 10 3. : Quant Question Archive [LOCKED]
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# If n is odd, then n(n*n - 1) is divisible by: 1. 9 2. 10 3.

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If n is odd, then n(n*n - 1) is divisible by: 1. 9 2. 10 3. [#permalink]

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22 Aug 2004, 08:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If n is odd, then n(n*n - 1) is divisible by:

1. 9
2. 10
3. 7
4. 24
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22 Aug 2004, 11:38
I pick D. 24.

n(n*n - 1) = n(n^2 - 1) = n(n-1)(n+1)
Since this is a product of 3 consecutive integers, the number is divisible by 3.

Moreover, An odd number can be expressed mathematically as 2k+1.

Here (2k+1)(2k)(2k+2) = 4*k*(k+1)*(2k+1) which means that the number is also divisible by 8. (k*(k+1) would be even)

3*8=24.
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22 Aug 2004, 11:50
The way I figured it is, 9 and 7 are automatically out as the number is atleast divisible by 3 and 2. That leaves 10 and 24. 10 is also out. Thus 24 has to be the answer.
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22 Aug 2004, 14:07
smcgrath12 wrote:
The way I figured it is, 9 and 7 are automatically out as the number is atleast divisible by 3 and 2. That leaves 10 and 24. 10 is also out. Thus 24 has to be the answer.

how did you eliminated 10?
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22 Aug 2004, 16:03
Anonymous wrote:

how did you eliminated 10?

The way I thought is, if 3 is a divisor, than 10 cannot be. Also if 10 is the divisor, than 5 has to be, and there is nothing to suggest 5 is.
22 Aug 2004, 16:03
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# If n is odd, then n(n*n - 1) is divisible by: 1. 9 2. 10 3.

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