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Number Properties. MGMAT CAT sum

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Source: — Data Sufficiency |
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by shovan85 » Tue Oct 19, 2010 1:33 am
As per B:

(k+1) / 3 is an odd number.

A general odd number can be written as 2n+1 where n is a number.

So (k+1) / 3 = 2n+1
=> k+1 = 6n+3
=> k = 6n + 2.

So the possible values of k = 2,8,14,...

When k = 2, (k+2) is divisible by 4 so the whole expression is div by 4
When k = 8, (K+2) is 10 divisible by 2 so the other part should be either div by 2 or 4 to get the whole expression as a multiple of 4. The other part (k^2+4K+3) if you put k = 8 it will be 99 not div by 2 or 4. So the whole expression is not div by 4.

Thus we can say Insufficient.
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by Geva@EconomistGMAT » Tue Oct 19, 2010 4:10 am
shovan85 wrote:As per B:

(k+1) / 3 is an odd number.

A general odd number can be written as 2n+1 where n is a number.

So (k+1) / 3 = 2n+1
=> k+1 = 6n+3
=> k = 6n + 2.

So the possible values of k = 2,8,14,...

When k = 2, (k+2) is divisible by 4 so the whole expression is div by 4
When k = 8, (K+2) is 10 divisible by 2 so the other part should be either div by 2 or 4 to get the whole expression as a multiple of 4. The other part (k^2+4K+3) if you put k = 8 it will be 99 not div by 2 or 4. So the whole expression is not div by 4.

Thus we can say Insufficient.
I used a similar approach with a more "qualitative" understanding, rather than resorting to 2n+1:
K+1/3 = odd integer so
k+1 = 3* odd integer
So k+1 is some odd multiple of 3, such as 3, 9, 15
k itself is "one less" than these multiples, or 2, 8, 14.

If k=2 then k+2 is divisible by 4, as Shovan said.
If k=8, then the previous statement should've already taught you that the expression is not divisible by 4: K+2 is 10, which is not divisible by 4, and the quadratic is composed of even+even+3 (since k itself is even), which will result with an odd number for any even k. So for k=8, the expression is not divisible by 4, leaving both a yes and a no answer - insufficient.
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by novel » Wed Oct 20, 2010 7:53 am
could some one please tell me how is 1 sufficient...
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by shovan85 » Wed Oct 20, 2010 8:03 am
novel wrote:could some one please tell me how is 1 sufficient...
k is div by 8 that means we can write k = 8x (where x is an integer).

Now put it in the Eqn: (k+2)*(k^2+4K+3) = (8x+2) * (64x^2 + 32x +3)

8x+2 is a multiple of 2 not 4. You can verify by putting some integer values of x/
x=0 8x+2 = 2
x=1 8x+2 = 10
.. ..
(64x^2 + 32x +3) the first two parts are clearly divisible by 4 as 64x^2 is a multiple of 4 and 32x is a multiple of 4.
So we can say 64x^2 + 32x is multiple of 4.

Now to a multiple of 4 if u add 3 then it will NEVER be a multiple of 4.

Thus sufficiently says 1 is not divisible by 4.

Hope this makes sense. Let me know if any further doubts.
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by novel » Wed Oct 20, 2010 9:01 am
shovan85 wrote:
novel wrote:could some one please tell me how is 1 sufficient...
k is div by 8 that means we can write k = 8x (where x is an integer).

Now put it in the Eqn: (k+2)*(k^2+4K+3) = (8x+2) * (64x^2 + 32x +3)

8x+2 is a multiple of 2 not 4. You can verify by putting some integer values of x/
x=0 8x+2 = 2
x=1 8x+2 = 10
.. ..
(64x^2 + 32x +3) the first two parts are clearly divisible by 4 as 64x^2 is a multiple of 4 and 32x is a multiple of 4.
So we can say 64x^2 + 32x is multiple of 4.

Now to a multiple of 4 if u add 3 then it will NEVER be a multiple of 4.

Thus sufficiently says 1 is not divisible by 4.

Hope this makes sense. Let me know if any further doubts.
thank you very much
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