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The integers r, s and t all have the same remainder when div
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Updated on: 23 Sep 2013, 04:58
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The integers r, s and t all have the same remainder when divided by 5. What is the value of t? (1) r + s = t (2) 20 <= t <= 24
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Originally posted by imhimanshu on 23 Sep 2013, 04:55.
Last edited by Bunuel on 23 Sep 2013, 04:58, edited 1 time in total.
Renamed the topic and edited the question.




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Re: The integers r, s and t all have the same remainder when div
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23 Sep 2013, 05:31




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Re: The integers r, s and t all have the same remainder when div
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26 Dec 2013, 21:50
rh410 wrote: I still don't understand how 1) r+s = t tells us that these must be multiples of 5. (And thus the remainder is 0)
Could someone please explain? Thanks! The question: when r,t,s are divided by 5, they yield the same remainder. Remember that when X is divided by 5, the remainder must be 0,1,2,3,4 ( remainder must always be smaller than divisor) If r and s has the same remainder, for example 1, then t can not have remainder 1 (21+11=32, remainder of the total (32) is the sum of remainders of 11 and 21, it equals 2 actually). So for the remainder to be the same, r,t and s must be multiples of 5 (20+10=30)




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Re: The integers r, s and t all have the same remainder when div
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23 Sep 2013, 06:32
imhimanshu wrote: The integers r, s and t all have the same remainder when divided by 5. What is the value of t?
(1) r + s = t (2) 20 <= t <= 24 For this r,s ant t can be any values which have same remainder when divided by 5 so r=6 s=1 t=21 or r=2, s=47 and t=1002 From statement 1: r + s = t => so for this only same remainder is "0". But r+s=t can be any value 5+10=15 where all the numbers have remainder "0" or 10+20=30 or etc . . .. So insufficient From statement 2: 20<=t<=24 so t can be 20, 21,22,23,24 Hence not sufficient From combining both the statements From 1 : r, s and t has to multiple of 5 so remainder is "0" From 2: t = 20,21,22,23,24 so from both the multiple of 5 is 20 t=20 sufficient hence both the statement are sufficient. answer is (c)



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Re: The integers r, s and t all have the same remainder when div
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26 Dec 2013, 20:34
I still don't understand how 1) r+s = t tells us that these must be multiples of 5. (And thus the remainder is 0)
Could someone please explain? Thanks!



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Re: The integers r, s and t all have the same remainder when div
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01 Dec 2014, 17:47
Bunuel wrote: (1) r + s = t > \((5a+x)+(5b+x)=5c+x\) > \(x=5(cab)\) . Can someone explain these steps? Thanks!



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02 Dec 2014, 01:21



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Re: The integers r, s and t all have the same remainder when div
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02 Dec 2014, 08:24
Bunuel wrote: JackSparr0w wrote: Bunuel wrote: (1) r + s = t > \((5a+x)+(5b+x)=5c+x\) > \(x=5(cab)\) . Can someone explain these steps? Thanks! \((5a+x)+(5b+x)=5c+x\); \(2x+5a+5b=5c+x\); \(x=5c5a5b\); \(x=5(cab)\). Hope it's clear. Thanks, I kept getting my signs mixed up...



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The integers r, s and t all have the same remainder when div
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12 Oct 2015, 10:11
I don't get how the answer to this is not B?
Between 20 and 24, 20 is the only number that can be comprised of numbers for which when divided by 5 give equal remainders:
Think about the numbers for t that fit in the inequality and any numbers below t that give the same remainder:
t= 20... 5, 10, 15, 20 t= 21... 6, 11, 16, 21 t= 22... 7, 12, 17, 22
and so on... The pattern becomes clear that only 20 works. What am i missing here?



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Re: The integers r, s and t all have the same remainder when div
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12 Oct 2015, 20:42
dubyap wrote: I don't get how the answer to this is not B?
Between 20 and 24, 20 is the only number that can be comprised of numbers for which when divided by 5 give equal remainders:
Think about the numbers for t that fit in the inequality and any numbers below t that give the same remainder:
t= 20... 5, 10, 15, 20 t= 21... 6, 11, 16, 21 t= 22... 7, 12, 17, 22
and so on... The pattern becomes clear that only 20 works. What am i missing here? You are using info from (1) when evaluating (2).
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Re: The integers r, s and t all have the same remainder when div
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06 Aug 2016, 14:17
Bunuel wrote: dubyap wrote: I don't get how the answer to this is not B?
Between 20 and 24, 20 is the only number that can be comprised of numbers for which when divided by 5 give equal remainders:
Think about the numbers for t that fit in the inequality and any numbers below t that give the same remainder:
t= 20... 5, 10, 15, 20 t= 21... 6, 11, 16, 21 t= 22... 7, 12, 17, 22
and so on... The pattern becomes clear that only 20 works. What am i missing here? You are using info from (1) when evaluating (2). Damn! I was so deep drilling (2) that by the time solved it I was sure I hadn't used info from (1). And I thought I was avoiding C trap! Thanks!



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Re: The integers r, s and t all have the same remainder when div
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29 Jan 2018, 16:46
here's the way i saw this question...
 since we're told r, s, and t all have the same remainder, it got me thinking: they would all have the same remainder if they were all multiples of 5.
(1) r+s=t r=10, s=10, t=20. works r=20, s=40, t=60. works ** we have 2 possible values for t, so insufficient.
(2) t is between 20 and 24 t can literally equal 20, 21, 22, 23 or 24...clearly insufficient
(3) we know t should be a multiple of 5 and within the range 2024. there's a # here (20) that fits these 2 criteria. sufficient.




Re: The integers r, s and t all have the same remainder when div &nbs
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29 Jan 2018, 16:46






