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The number properties

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Source: — Data Sufficiency |
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by eagleeye » Tue Jun 26, 2012 8:32 am
Hi nidhis.1408:

We are given that y and n are positive integers, and need to find whether y*n is divisible by 7.

Let's look at the statements:

(1)n^2- 14n+49=0
This reduces to n^2- 14n+49 = (n-7)^2.
Since n^2- 14n+49=0, n-7=0 so n=7. Hence yn is divisible. sufficient.

(2) n + 2 is the first of three consecutive integers
whose product is 990.
Here we have (n+2)*(n+3)*(n+4) = 990. A quick breakdown of 990 reveals that 990 = 9*10*11.
Hence (n+2)*(n+3)*(n+4) = 9*10*11. Hence n+2=9 or n=7. Again sufficient.

The final answer is, therefore, D.

Let me know if this helps :)
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by nidhis.1408 » Tue Jun 26, 2012 8:36 am
I forgot to do n+2 = 9. i took n as 9!
thanks a lot!
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by Brent@GMATPrepNow » Tue Jun 26, 2012 9:58 am
See below...

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Thu Jun 28, 2012 6:17 am, edited 1 time in total.
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by hey_thr67 » Wed Jun 27, 2012 6:39 pm
, if the product of 3 consecutive integers is 990, then there can be only 3 consecutive integers that satisfy this condition. In other words, there is only one possible value for the first of those three consecutive integers.
Brent, you seem to be doing generalization here. for what types of number it is true ?
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by Brent@GMATPrepNow » Thu Jun 28, 2012 6:17 am
hey_thr67 wrote:Brent, you seem to be doing generalization here. for what types of number it is true ?
Sorry, I need to retract what I said earlier (see edit).

If we know that n is divisible by 7, then we can be certain that yn is divisible by 7.
Since 9x10x11=990 (from statement 2), we know that n+2=9, which means n=7, in which case yn must be divisible by 7.

Cheers,
Brent
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