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

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Re: The integers r, s and t all have the same remainder when div [#permalink]
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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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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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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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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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The integers r, s and t all have the same remainder when div [#permalink]
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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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]
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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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 20-24. there's a # here (20) that fits these 2 criteria. sufficient.
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Re: The integers r, s and t all have the same remainder when div [#permalink]
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Hi All,

We're told that The R, S and T are all INTEGERS and that all have the SAME remainder when divided by 5. We're asked for the value of T. This question can be solved in a couple of different ways - and it's built around a math pattern that you might not realize is there (unless you TEST VALUES to prove it).

To start, for 3 numbers to have the SAME remainder when divided by 5, they will fit a particular pattern:
IF....
the remainder is 0, then the numbers will be evenly divisibly by 5 --> such as 0, 5, 10, 15, 20, 25, etc
the remainder is 1, then the numbers will all be "1 more" than a multiple of 5 --> such as 1, 6, 11, 16, 21, 26, etc
the remainder is 2, then the numbers will all be "2 more" than a multiple of 5 --> such as 2, 7, 12, 17, 22, 27, etc
Etc.

(1) R + S = T

Using the above patterns, if we add two numbers with the SAME remainder, then we'll end up with the sum of their two remainders. For example, if R= 6 and S= 11, then we'll have a sum of 17... but 17 has a remainder of TWO (re: the sum of the two individual remainders of 1) when it's divided by 5. We're told that all 3 integers must have the SAME remainder though, so R and S clearly CANNOT have remainders of 1.

This issue occurs with all of the other options EXCEPT for remainders of 0.... for example, if R=5 and S=10, then we'll have a sum of 15... and 15 ALSO has a remainder of 0. This ultimately means that R, S and T MUST each be a multiple of 5, but we don't know exactly which multiple of 5 T is.
Fact 1 is INSUFFICIENT

(2) 20 <= T <= 24

Fact 2 gives us 5 possible values for T: 20, 21, 22, 23 and 24.... but we don't know which one it is (and we have no additional information that relates the values of R and S to the value of T, so there's no way to define the exact value of T.
Fact 2 is INSUFFICIENT

Combined, we know...
T must be a multiple of 5.
T is one of the numbers from 20 to 24, inclusive.

The only value that 'fits' both pieces of information is 20.
Combined, SUFFICIENT

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Re: The integers r, s, and t all have the same remainder when divided by 5 [#permalink]
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Bunuel 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

Given: The integers r, s, and t all have the same remainder when divided by 5

I.e. r = 5a+p
I.e. s = 5b+p
I.e. t = 5c+p

where p is the remainder.

Question: t = t = 5c+p = ?

Statement 1: r + s = t

I.e. (5a+p) + (5b+p) = 5c+p

I.e. 5(a+b) + p = 5c

I.e. p must be zero and t must be a multiple of 5

but c is unknown hence value of t can not be calculated

NOT SUFFICIENT

Statement 2: 20 ≤ t ≤ 24

t can be 20 or 21 or 22 etc hence

NOT SUFFICIENT

Combining the statements

t is a multiple of 5 and 20 ≤ t ≤ 24

I.e. t = 20

SUFFICIENT

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Re: The integers r, s, and t all have the same remainder when divided by 5 [#permalink]
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Bunuel 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

r/5=5p+x
s/5=5q+x
t/5=5r+x

0 ≤ remainder x ≤ 4

(1) insufic

5p+x+5q+x=5r+x
5p+5q+x=5r
only multi of 5 for x is 0
5(p+q)=5r

(2) insufic

(1/2) sufic

20≤5r+x≤24
20≤5r≤24
4≤r<5
r=integer=4

Ans (C)
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Re: The integers r, s and t all have the same remainder when div [#permalink]
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When r, s and t are divided by 5, they all leave a common remainder, say L. The value of t needs to be found.
A quick look at the statement reveals that statement 2 is easier to interpret, so let’s consider statement 2 alone.

From statement 2 alone, we gather that t can take any of the five values among {20, 21,22, 23 ,24}. Clearly, no unique value of t.
Statement 2 alone is insufficient. Answer options B and D can be eliminated. Possible answer options are A, C or E.

From statement I alone, r + s = t.
From question data, we know that r, s and t, all leave the same remainder when divided by 5 (which we have considered as L).

Therefore, r = 5x + L, s = 5y + L and t = 5k + L. Substituting these values in the equation r+s = t, we have,

5x + L + 5y + L = 5k + L.

Simplifying, we have k = (x+y) + $$\frac{L}{5}$$. We have only one equation with more than one unknowns in it.

Statement I alone is insufficient. Answer option A can be eliminated. Possible answer options are C or E.

Combining statements I and II, we have the following:
From statement II, we have a finite set of values for t i.e. 20, 21, 22, 23 and 24. The remainders when each of these values are divided by 5 are 0, 1, 2, 3 and 4.

Now, when r and s are divided by 5, they should also leave the same remainders in the same order. That’s only possible if t = 20.

If t = 20, r + s = 20 can be satisfied by many combinations of r and s such that they leave 0 as the remainder when divided by 0.

On the other hand, if t = 21, the remainder L = 1. If r = 11, then s = 10 and both do not leave the same remainders.

This happens because of the innate nature of the equation 5x + L + 5y + L = 5k + L. Only when L = 0 can the two sides be equal. For all other values of L, there will be an extra value on the LHS which means the equation will not hold.

The combination of statements is sufficient to find the value of t as 20. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!
Aravind B T
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Re: The integers r, s and t all have the same remainder when div [#permalink]
Correct me if im wrong but:

7 and 17 both have remainder of 2 and sums up to 24.
6 and 16 both have remainder of 1 and sums up to 22.

What am i not getting?

Thanks!
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Re: The integers r, s and t all have the same remainder when div [#permalink]
thommyGmate wrote:
Correct me if im wrong but:

7 and 17 both have remainder of 2 and sums up to 24.
6 and 16 both have remainder of 1 and sums up to 22.

What am i not getting?

Thanks!

Hi thommyGmate,

The prompt tells us that all 3 numbers (R, S and T) are all INTEGERS and that all have the SAME remainder when divided by 5. We're asked for the value of T.

Your two examples do NOT fit what we're told though. Here's why:

While 7 and 17 both have a remainder of 2 (when dividied by 5), the number 24 does NOT (it has a remainder of 4).
While 6 and 16 both have a remainder of 1 (when dividied by 5), the number 22 does NOT (it has a remainder of 2).

In my explanation (above), you'll see how there actually IS a pattern that you can take advantage of to answer this question.

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Rich

Contact Rich at: Rich.C@empowergmat.com
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