In need of a quicker method in solving this....

smclean23

If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?
(1) N is divisible by 3.
(2) N is divisible by 7.

THANKS!

Vignesh.4384 · Posted: Sat Jul 05, 2008 8:40 pm Post subject:

Answer is B
I am not sure of a faster method 2 do this. But jus think like this.

Given statement: N = 4,321 + K

Clue : K is a positive integer less than 10.

when K is a +ve no less than 10 (given)

N can be one of these possible vales {4322 or 4323 or 4324 or 4325 or 4326 or 4327 or 4328 or 4329 or 4330}.

From option A

Surely u cannot determine the unique value of K because, as u see above N is set of 9 nos and the vale of K can vary(simple reason between 9 consective nos, at the max 3 nos wil be a multiple of 3 and min of 2 no will be mltiples of 3).

From option B

We can uniquely determine the value of K because in a set of 9 consecutive nos,the multiple of 7 can occur at the max 2 times. But since in the set N 4326 is perfectly divisible by 7 so it means the next no divisible by 7 wil be
4333. however since 4333 is outside set N, value of K can be determined uniqely

Believe me .. dosent take more than 1 min to do this problem using this method.

I dont know of a qicker method .. If anybody can think of another way pls let us know.

Regards.
Vignesh

ed09 · Posted: Sat Jul 05, 2008 11:14 pm Post subject:

My way of reasoning is:

(1) by criterion of divisibility by 3
4323 (k=2), 4326 (k=5), 4329 (k=8) - all these numbers are divisible by 3
=> ambiguity of definite and unique k => INSUFFICIENT

(2) Let's look at the given number as a sum of summands each of which is divisible by 7.
Take them simple!
4,320 = 4200 + 70 + 35 + (16)
Evidently, all of the summands but 16 are divisible by 7.
Hence if k=5 than 16+5=21 => the sum (N) will be divisible by 7.
The next k (k>0) could be only if k=k1+7=5+7=12, but this is not allowed by the main clause.
Thus if N is divisible by 7, k=5 is the only acceptible variant.
SUFFICIENT

The answer is B.

Best!

Ian Stewart · Posted: Sun Jul 06, 2008 7:53 am Post subject:

Good approaches to the problem above. Just wanted to correct one small thing:

Vignesh.4384 · Posted: Sun Jul 06, 2008 8:00 am Post subject:

Hi Ian,

Thanks for correcting me.
I guess i was a bit careless when i said that

(simple reason between 9 consective nos, at the max 3 nos wil be a multiple of 3 and min of 2 no will be mltiples of 3).

It should hav been

(simple reason between 9 consective nos, at the max 4 nos wil be a multiple of 3 and min of 3 no will be mltiples of 3).

Regards,
Vignesh

wilderness · Posted: Mon Jul 07, 2008 1:44 pm Post subject:

Hi Vignesh,

I have to side with Ian here, because between 9 consecutive numbers there will be always 3 and only 3 multiples of 3. I Could not figure out a set of 9 consecutive numbers that has 4 multiples of 3 in it. Can you throw some light.

Regards,

Ian Stewart · Posted: Mon Jul 07, 2008 5:02 pm Post subject:

target790 · Posted: Thu Jul 10, 2008 11:01 am Post subject:

K=(1 ..9)

(1)A number is divisible by 3 when the sum of it's digits is divisible by 3

So in this case k can be 2,3,6

Not sufficient

(2)7 being prime number k can only be 5

Sufficient.

So B

Enjoy Smile