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If \(X\) is a positive integer, what is the remainder of \(\frac{X}{8}\)? (1) The remainder of \(\frac{X}{16}\) is 2. (2) The remainder of \(\frac{X}{24}\) is 10.
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16 Sep 2014, 00:41
Official Solution: Statement (1) by itself is sufficient. From S1 it follows that \(X = 16N + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2. Statement (2) by itself is sufficient. From S2 it follows that \(X = 24N + 10 = (24N + 8) + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2. Answer: D
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09 Nov 2014, 08:07
I'd appreciate if you could explain what you did for st 2.. how did you get x/8 ?



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10 Dec 2014, 15:24
Split 24N + 10 ==> (24N+8) + 2
It can further be written as 8(3N+1) + 2. When this is divided by 8 the remainder is 2.



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Re: M1012
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23 Feb 2016, 00:48
Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From S1 it follows that \(X = 16N + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2. Statement (2) by itself is sufficient. From S2 it follows that \(X = 24N + 10 = (24N + 8) + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2.
Answer: D Hi I think I am missing some concept here. I framed equations like this: Stem=X=8a+R; To find R=X8a Stmt1=X=16b+2 Stmt2=X=24c+10 Hence got E as cant find a or X from any of the stmts. I knw logically what I did is not wrong but dont know how to proceed from here. Read something about primes but how does the concept come about? Pls elaborate. Thanks.



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sinhap07 wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is sufficient. From S1 it follows that \(X = 16N + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2. Statement (2) by itself is sufficient. From S2 it follows that \(X = 24N + 10 = (24N + 8) + 2\) where \(N\) is an integer. Thus, the remainder of \(\frac{X}{8}\) is 2.
Answer: D Hi I think I am missing some concept here. I framed equations like this: Stem=X=8a+R; To find R=X8a Stmt1=X=16b+2 Stmt2=X=24c+10 Hence got E as cant find a or X from any of the stmts. I knw logically what I did is not wrong but dont know how to proceed from here. Read something about primes but how does the concept come about? Pls elaborate. Thanks. Hi, NOTE: whatever is the remainder by an integer, the remainder will remain teh same if div by its factors.. only thing to check is if the R is further div by that factor what you should have done after making your statements.. Stmt1=X=16b+2 we do not require X or a, we are interested in R.. so what does x/8 mean.. (16b+2)/8=16b/8 + 2/8.. 16b is div by 8 so will not leave any remainder but 2/8 will leave a remainder of 2.. suff
Stmt2=X=24c+10 same way ( 24c+10)/8= 3c +10/8.. when 10 is div by 8, it will give us 2 as remainder .. again suff
D
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Re: M1012
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03 May 2016, 06:09
could someone explain to me how does it follow from statement 1 that
that X=16N+2X=16N+2 where N is an integer?



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Re: M1012
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03 May 2016, 06:22
nausherwan wrote: could someone explain to me how does it follow from statement 1 that
that X=16N+2X=16N+2 where N is an integer? hi I. (1) The remainder of \(\frac{X}{16}\) is 2... it can be written as  when x is divided by 16, the remainder is 2... we do not know the quotient(Q) here but just know the remainder.. how do I write it in a equation x = 16a + 2.. a is the quotient and 2 is the remainder.. say a is 5.. so x= 5*16 +2 = 82.. I can write it as the remainder of 82/16 is 2.. so x=16n+2 is just another way of writing  The remainder of \(\frac{X}{16}\) is 2
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Re: M1012
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03 May 2016, 06:24
nausherwan wrote: could someone explain to me how does it follow from statement 1 that
that X=16N+2X=16N+2 where N is an integer? Check the links below: Divisibility and Remainders on the GMATTheory on remainders problemsTips on remaindersCyclicity on the GMATUnits digits, exponents, remainders problemsHope it helps.
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Re: M1012
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14 Nov 2016, 22:58
why did you consider N as an integer.it was not given in the question?



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Re: M1012
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27 May 2017, 10:08
Another approach  Write out few of the Numbers and check S1: The Remainder of \(\frac{X}{16}= 2\) This means X = 2, 18, 34... Remainders when divided by 8 in all the cases = 2 Suff S2: The Remainder of \(\frac{X}{24}= 10\) This means X = 10, 34, 58... Remainder when X is divided by 8 in all cases = 2 Suff Final answer : D
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Re M1012
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31 Jan 2018, 00:53
I think this is a highquality question and I don't agree with the explanation. what if we take x=14 for statement 1. then 14/16 gives rem as 2 but 14/8 gives rem as 6. so the correct answer should be B.



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31 Jan 2018, 01:49
muditsingh0616 wrote: I think this is a highquality question and I don't agree with the explanation. what if we take x=14 for statement 1. then 14/16 gives rem as 2 but 14/8 gives rem as 6. so the correct answer should be B. 14 divided by 16 gives the remainder of 14, not 2. From (1) x could be 2, 18, 34, 50, 66, ... Each of these values gives a remainder of 2 when divided by 8, so (1) is sufficient.
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Re: M1012
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04 Nov 2018, 11:05
Don't see how S) 1 is sufficient.
x/16 gives remainder 2.
Hence X = 16q + 2
if q=0, x= 16(0) + 2 =2
if q=1 x= 16 + 2 = 18
lets take x=2. 2/8 gives 0 remainder 8. lets take x=18 18/8 gives 2 remainder 2.
Insufficient.
S) 2 on the other hand..
X = 24p + 10
if p=0 x= 10
p=1 x = 34
10/8 gives remainder 2 34/8 gives remainder 2.
Statement 2 works, but 1 doesn't.



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Re: M1012
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08 Mar 2019, 13:59
PierTotti17 wrote: Don't see how S) 1 is sufficient.
x/16 gives remainder 2.
Hence X = 16q + 2
if q=0, x= 16(0) + 2 =2
if q=1 x= 16 + 2 = 18
lets take x=2. 2/8 gives 0 remainder 8. lets take x=18 18/8 gives 2 remainder 2.
Insufficient.
S) 2 on the other hand..
X = 24p + 10
if p=0 x= 10
p=1 x = 34
10/8 gives remainder 2 34/8 gives remainder 2.
Statement 2 works, but 1 doesn't. PierTotti17 All is good in what you wrote, except, 2/8 gives a remainder of 2, not 8. Hence 1 is sufficient as well.



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Re: M1012
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25 Aug 2019, 08:12
Easy to prove sufficiency with number picking, just plug in 2, 18 for St 1) you will get remainder 2 both times when dividing by 8. St 2) must give the same remainder... you get 2 when dividing by 8 as well, plug in 2 and 34. This is because both 16 and 24 are multiples of 8 so they all share the same remainder when dividing X. As chetan2u mentioned the only thing is to make sure remainder/divisor is in the most reduced form.



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Re: M1012
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29 Nov 2019, 01:37
My two cents' worth here, probably a rehash of what the experts have expounded on above:
In both statements, the divisors 16, 24 (in relation and including the divisor '8' in the stem) are all multiples of the smallest of the divisors: '8'.
If so, we can possibly infer that all outcomes when divided by the divisors (=8K) would yield the same remainder; without calculating, I would pick 'D'.
Here are some numbers to try out e.g. 2,4,8 as divisors; the same numerator X=17 is tested. 17/2 with R=1 17/4 with R=1 17/8 with R=1
Each statement is sufficient on its own, hence D.










