Remainder problem

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Remainder problem

by raunekk » Sat May 21, 2011 11:06 pm
Q) When a number is divided by 13, the remainder is 11. When the same number is divided by 17, the remainder is 9. What is the number?

1) 339
2)349
3) 359
4) 369
5) data inadequate

What is the shortest possible method to solve such sums.

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by Frankenstein » Sun May 22, 2011 2:12 am
Hi,
In such questions, it is better to go from options. By checking the options we find that 349 satisfies both conditions. But, there will be many numbers which satisfy this condition.
All such numbers will be of the form 349+(LCM of 13,17).(n) where n is an integer.
So, the answer is E

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by peelamedu » Fri Jun 10, 2011 4:46 am
I didn't quite get this.
Could experts help ?

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by newgmattest » Fri Jun 10, 2011 4:54 am
Hi GMAT Experts,

I am bit confused with this answer as similar to this there was one median question and there were many possibility, but one option choice was same as one value and we picked that instead of "none". Why here we don't pick number as that is one option?

What is exact logic behind answering such question(I believe it had been DS question, then definitely there will be many numbers), but as PS, we must choose answer if it is available rather than picking "insufficient".

Please help to clarify.

Thanks a lot.

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by Frankenstein » Fri Jun 10, 2011 5:32 am
Hi,
If the question is : What is the number?
If there are many possible values and one of them is in options, and if it is a real GMAT question, you will definitely have an option like 'data inadequate'.

If the question is : The number can be ?
can be - likely not certainty
So, you can choose one of the values in the options that satisfies the conditions.
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by cans » Fri Jun 10, 2011 6:00 am
a)339 when divided by 13, remainder=1
b)349 when divided by 13, remainder=11; when divided by 17, remainder=9
IMO B
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by finites » Fri Jun 10, 2011 11:03 am
plugging the answers will solve it..

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by breakkgmat » Sat Jun 11, 2011 9:50 am
I would solve in this way..Let the number is X & Quotient is Y..
So,if first case. X=13Y+11
& X=17Y+9

if you solve this, 17Y+9=13Y+11 or 4Y = 2 OR Y=1/2 WHICH IS NOT AN INTERGER,
Putting the value of Y in equation Y we get..
X=13*1/2 +11
S0, X is not an integer..

Ans.E.

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by Frankenstein » Sat Jun 11, 2011 10:03 am
breakkgmat wrote:I would solve in this way..Let the number is X & Quotient is Y..
So,if first case. X=13Y+11
& X=17Y+9

if you solve this, 17Y+9=13Y+11 or 4Y = 2 OR Y=1/2 WHICH IS NOT AN INTERGER,
Putting the value of Y in equation Y we get..
X=13*1/2 +11
S0, X is not an integer..

Ans.E.
Hi,
You have taken for granted that the quotient is same when X is divided by 13 and 17, which is not the case. Moreover, you can't have fraction as quotient.
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by smackmartine » Sat Jun 11, 2011 10:26 am
IMO B
One alternative approach could be focusing on "number is divided by 13, the remainder is 11"

subtract 11 from all options in order to check their divisibility by 13. Its B (349-11)= 338.
Taking this option , work with 17
340 is 20 time 17 and 349 leaves 9 reminder on dividing it by 17.

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by Frankenstein » Sat Jun 11, 2011 10:31 am
smackmartine wrote:IMO B
One alternative approach could be focusing on "number is divided by 13, the remainder is 11"

subtract 11 from all options in order to check their divisibility by 13. Its B (349-11)= 338.
Taking this option , work with 17
340 is 20 time 17 and 349 leaves 9 reminder on dividing it by 17.
Hi,
I agree that 349 satisfies the condition. But, there can be many other numbers. So, to say that '349 is the number', I believe we need more information. So, I prefer E.
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by smackmartine » Sat Jun 11, 2011 1:16 pm
Frankenstein wrote:
smackmartine wrote:IMO B
One alternative approach could be focusing on "number is divided by 13, the remainder is 11"

subtract 11 from all options in order to check their divisibility by 13. Its B (349-11)= 338.
Taking this option , work with 17
340 is 20 time 17 and 349 leaves 9 reminder on dividing it by 17.
Hi,
I agree that 349 satisfies the condition. But, there can be many other numbers. So, to say that '349 is the number', I believe we need more information. So, I prefer E.
IMO unless we prove that rest of the options cannot be obtained using the information in the question, we cannot select option E.While this reasoning holds true only for problem solving questions, if this were Data sufficiency Question, I would agree with you, because there can be other possibilities.

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by Ian Stewart » Sat Jun 11, 2011 4:10 pm
If I ask the question:

x is an even integer. What is the value of x?
A) 1
B) 2
C) 3
D) 5
E) cannot be determined


it should be clear that the answer is E here; we don't have enough information to find the value of x. The same thing is true about the question in the original post. When you know the remainder when x is divided by 13 and by 17, the possible values of x will be separated by the LCM of 13 and 17, which is 221. That is, there will be an infinite number of possible values of x here, and these values will be 221 apart. So yes, it's possible that x = 349, but it's equally possible that x = 349 - 221 = 128, or that x = 349 + 221 = 570, or that x = 570 + 221 = 791, among many other possibilities.

I'd add that the question is poorly presented (the GMAT would never have an answer choice which read "data inadequate", for one thing), which I imagine has contributed to some of the confusion about the correct answer.
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by smackmartine » Sat Jun 11, 2011 4:45 pm
Ian, I see your and Frankenstein's point. However, I will try to find a similar PS question, which found couple of months back.

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