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# what will be remainder when x is divided by 31? 1. x + 10 /

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what will be remainder when x is divided by 31? 1. x + 10 / [#permalink]

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15 Aug 2010, 06:38
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what will be remainder when x is divided by 31?

1. x + 10 / 31 has no remainders.
2. x + 33 / 10 has no remainders.
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15 Aug 2010, 08:58
Statement 1 - possible values of "X" = 21, 52, 83, 114 ...... all giving different remainders when divided by 31
NOT SUFFICIENT - options A & D are ruled out

Statement - 2 - possible values of "X" = 7,17,27,47,57,67,107........................ giving different values of remainder when divided by 31 => NOT SUFFICIENT - option B is out.

Combining them = X should have units digit 1 (statement 2) & must be 10 lesser than any multiple of 31 (statement 1)
Multiple of 31 having last digit 7 = 31X7 = 217 = X+10 (from statement 1) => X= 207 & remainder for 207 divided by 31 = 21
Next multiple of 31 having last unit digit 7 is 31X17 = 527 = X+10 => X = 517 & remainder when divided by 31 = 21

21 is the remainder for all the cases satisfying both conditions.
Please let me know whats the OA and whether I am correct.
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15 Aug 2010, 09:15
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Expert's post
efet wrote:
what will be remainder when x is divided by 31?

1. x + 10 / 31 has no remainders.
2. x + 33 / 10 has no remainders.

Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So most likely GMAT would present the above question is the following way:

What is the remainder when the positive integer x is divided by 31?

(1) When x+10 is divided by 31 the remainder is zero.
(2) When x+33 is divided by 10 the remainder is zero.

Note that the questions are basically the same and have the same answers, but only the second one is GMAT type.

Now, we can solve this question algebraically or with number plugging, which in this case is easier:

(1) When x+10 is divided by 31 the remainder is zero --> x+10 to be multiple of 31 x must be 21, 52, 83, ... each yields a remainder of 21 upon division by 31. Sufficient.

(2) When x+33 is divided by 10 the remainder is zero --> x+33 to be multiple of 10 x must be 7, 17, 27, ... 7 yields a remainder of 7 upon division by 31, but 17 yields a remainder of 17 upon division by 31. Not sufficient.

Hope it helps.
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15 Aug 2010, 10:21
Sorry!!
I considered 21 to be one of the possibilities which comes from option 1.
If we don't consider 21 as an option then answer is A.
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15 Aug 2010, 10:40
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soumanag wrote:
Sorry!!
I considered 21 to be one of the possibilities which comes from option 1.
If we don't consider 21 as an option then answer is A.

21 is valid value for x in statement (1) and 21 divided by 31 also gives remainder of 21.

Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0\leq{r}<d$$ (remainder is non-negative integer and always less than divisor).

So when divisor (31 in our case) is more than dividend (21 in our case) then the reminder equals to the dividend:

21 divided by 31 yields a reminder of 21 --> $$21=0*31+21$$.

Hope it helps.
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15 Aug 2010, 11:01
Crystal clear!! thanks
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15 Aug 2010, 18:26
what will be remainder when x is divided by 31?

1. x + 10 / 31 has no remainders.
2. x + 33 / 10 has no remainders.

Positive integer a divided by positive integer d yields a reminder of r can always be expressed as a=qd+r

from stat 1 x+10=31Q
not sufficient
from stat 2 x+33=10Q
not sufficient
now consider both stat 1 & 2
therefore c
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16 Aug 2010, 03:48
anilnandyala wrote:
what will be remainder when x is divided by 31?

1. x + 10 / 31 has no remainders.
2. x + 33 / 10 has no remainders.

Positive integer a divided by positive integer d yields a reminder of r can always be expressed as a=qd+r

from stat 1 x+10=31Q
not sufficient
from stat 2 x+33=10Q
not sufficient
now consider both stat 1 & 2
therefore c

Statement (1) is sufficient, so the answer to this question is A, not C. Number plugging method solution is in my previous post, below is algebraic approach for statement (1).

(1) When x+10 is divided by 31 the remainder is zero --> $$x+10=31q$$ --> $$x=31q-10$$ --> $$x=31q-31+21$$ --> $$x=31(q-1)+21$$ directly tells us that remainder upon division $$x$$ by 31 is 21. Sufficient.

Hope it's clear.
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16 Aug 2010, 08:09
Great explanation.
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16 Aug 2010, 16:19
Thanks for the solution Brunel!
This is a quick and easy solution

Will your method also work for question like this that will need both equations (ie. C) to solve?
Re: remainder   [#permalink] 16 Aug 2010, 16:19
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