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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?

\(r=5a+x\) \(s=5b+x\) \(t=5c+x\)

Where the remainder x is \(0\leq{x}<5\).

(1) r + s = t --> \((5a+x)+(5b+x)=5c+x\) --> \(x=5(c-a-b)\) --> the remainder (x) is a multiple of 5, thus it's 0 --> r, s and t are multiples of 5. Not sufficient.

(2) 20 <= t <= 24. Clearly insufficient.

(1)+(2) There is only one multiple of 5 between 20 and 24, inclusive, namely 20. Sufficient.

Re: The integers r, s and t all have the same remainder when div [#permalink]

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23 Sep 2013, 07:32

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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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26 Dec 2013, 22:50

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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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16 Aug 2017, 10:20

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