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
In need of a quicker method in solving this....
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Source: Beat The GMAT — Data Sufficiency |
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Vignesh.4384
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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!
(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!
GMAT/MBA Expert
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Ian Stewart
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Good approaches to the problem above. Just wanted to correct one small thing:
If you have nine consecutive integers, it will always be true that exactly three of them are divisible by 3.Vignesh.4384 wrote: (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).
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Vignesh.4384
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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
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
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wilderness
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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,
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,
GMAT/MBA Expert
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Ian Stewart
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Yes, that's exactly right. If you have nine consecutive integers, exactly three will be multiples of 3, exactly three will give a remainder of 1 when you divide by three, and exactly three will give a remainder of 2 when you divide by three. It's not possible for there to be four multiples of three among nine consecutive integers.wilderness wrote: between 9 consecutive numbers there will be always 3 and only 3 multiples of 3.
