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# The integers r, s and t all have the same remainder when div

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

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23 Sep 2013, 05:55
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Sep 2013, 05:58, 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 [#permalink]

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23 Sep 2013, 06:31
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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.

Hope it's clear.
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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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Re: The integers r, s and t all have the same remainder when div [#permalink]

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26 Dec 2013, 21:34
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)

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

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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)

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 [#permalink]

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01 Dec 2014, 18:47
Bunuel wrote:
(1) r + s = t --> $$(5a+x)+(5b+x)=5c+x$$ --> $$x=5(c-a-b)$$ .

Can someone explain these steps?

Thanks!

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

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02 Dec 2014, 02:21
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JackSparr0w wrote:
Bunuel wrote:
(1) r + s = t --> $$(5a+x)+(5b+x)=5c+x$$ --> $$x=5(c-a-b)$$ .

Can someone explain these steps?

Thanks!

$$(5a+x)+(5b+x)=5c+x$$;

$$2x+5a+5b=5c+x$$;

$$x=5c-5a-5b$$;

$$x=5(c-a-b)$$.

Hope it's clear.
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Re: The integers r, s and t all have the same remainder when div [#permalink]

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02 Dec 2014, 09:24
Bunuel wrote:
JackSparr0w wrote:
Bunuel wrote:
(1) r + s = t --> $$(5a+x)+(5b+x)=5c+x$$ --> $$x=5(c-a-b)$$ .

Can someone explain these steps?

Thanks!

$$(5a+x)+(5b+x)=5c+x$$;

$$2x+5a+5b=5c+x$$;

$$x=5c-5a-5b$$;

$$x=5(c-a-b)$$.

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 [#permalink]

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12 Oct 2015, 11:11
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 [#permalink]

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12 Oct 2015, 21:42
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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 [#permalink]

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06 Aug 2016, 15:17
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 [#permalink]

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16 Aug 2017, 10:20
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Re: The integers r, s and t all have the same remainder when div   [#permalink] 16 Aug 2017, 10:20
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