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# When the positive integer n is divided by 25, the remainder

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When the positive integer n is divided by 25, the remainder [#permalink]

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15 Sep 2007, 11:19
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When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

(1) n < 100
(2) When n is divided by 20, the remainder is 3.

[Reveal] Spoiler: My take
it is c, and i got 63 as n using both statements, but wasn't certain there didn't exist another one.

any quick way to do this.

thx
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15 Sep 2007, 11:55
When the positive integer n is divided by 25, the remainder is 13.

n = 25*x+13

statement 1

n < 100

n = 13,38,63,88

insufficient

statement 2

when n is divided by 20, the remainder is 3.

n = 20*y + 3

20*y + 3 = 25*x+13

insufficient

both statements

n = 13,38,63,88

n = 20*y + 3 ---> 3,23,43,63,83

n = 63

sufficient

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03 Feb 2008, 13:07
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When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

1. n < 100
2. When n is divided by 20, the remainder is 3.
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03 Feb 2008, 14:22
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Expert's post
C

1. 38,63,88. insuff.
2. 63,163. insuff.

1&2. 63. suff.
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03 Feb 2008, 14:26
I tried solving this via the remainder formula. Is their anyway to do this without picking #s?

I got as far as:

n = 25k + 13
n = 20k +3

No idea if I was on the right track with this one. I hate remainder problems.
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03 Feb 2008, 14:29
I also hate remainder problems
Smart number approach seems to be faster way.
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03 Feb 2008, 14:57
n = 25k + 13
n = 20m +3

25k + 13 = 20m +3
25k + 10 = 20m
5k + 2 = 4m
m=(5k+2)/4 - m has to be integer.
k has to be even and not divisible by 4

k=2,6,10 ==> 63, 163, 263 - our magic integers
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03 Feb 2008, 17:39
I missing you at this part:

m=(5k+2)/4 - m has to be integer.
k has to be even and not divisible by 4

Why does k have to be even, and not divisible by 4?
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03 Feb 2008, 23:43
jimmyjamesdonkey wrote:
I missing you at this part:

m=(5k+2)/4 - m has to be integer.
k has to be even and not divisible by 4

Why does k have to be even, and not divisible by 4?

1. m has to be an integer.
2. (5k+2) has to be even and divisible by 4
3. (5k+2) is even when 5k is even. Therefore k is even.
4. if 5k is divisible by 4, (5k+2) will not divisible by 4: 5*4i+2=4*(5i)+2. Therefore, (5k+2) has not to be divisible by 4 => 5k has not to be divisible by 4
5. k is even and indivisible by 4.
6: 2,6,10....
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04 Feb 2008, 03:31
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My approach is :

n=25x+13

multiples of 25 are -> 25,50,75 thus n would be ->38,63,88

Apply 1-> I have 3 values -> Not sufficient
Apply 2-> I will have one value ->63, there could be more above 100, thus applying both 1&2
you get n=63
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Re: DS GMAT prep Remainder [#permalink]

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04 Sep 2011, 02:47
per ques stem, we can get values ---> 13,38,63,88,113 etc so not a single value of n is derived..

per stat 1 , we dnt get to knw anything..
per stat 2, we can get values ---> 3,23,43,63,83,103 etc

Now, there is only one value (63) which satisfies both the ques stenm condition and stat 2 condition..

somebody pls help me understand what am i missing to get the answer...
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Re: DS GMAT prep Remainder [#permalink]

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04 Sep 2011, 03:15
DeeptiM wrote:
per ques stem, we can get values ---> 13,38,63,88,113 etc so not a single value of n is derived..

per stat 1 , we dnt get to knw anything..
per stat 2, we can get values ---> 3,23,43,63,83,103 etc

Now, there is only one value (63) which satisfies both the ques stenm condition and stat 2 condition..

somebody pls help me understand what am i missing to get the answer...

St1 & Stem:
n can be: 13, 38, 63, 88
Not Sufficient.

St2 & Stem:
n can be: 63, 163, 263...
Not Sufficient.

Using both, we have just one integer in common "63". 63 must be the value because it satisfies both statements and the stem.
Sufficient.

Ans: "C"
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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25 Jan 2013, 23:58
I used the number pluggin approach until I got this no: 63.
When I was about to hit B as the answer, I saw the option A ..
Highly unlikely that A couldn't have been there for a reason. .

thought that there must be numbers greater than 100 which will satisfy B.
So Chose C :D
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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26 Jan 2013, 04:38
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When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that $$n=25q+13$$, so n could be 13, 38, 63, 88, 113, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> $$n=20p+3$$. n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.

(1)+(2) The only value of n which both statements is 63. Sufficient.

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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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06 Oct 2013, 01:55
Bunuel wrote:
When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that $$n=25q+13$$, so n could be 13, 38, 63, 88, 113, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> $$n=20p+3$$. n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.

(1)+(2) The only value of n which both statements is 63. Sufficient.

Out of the highlighted values only 63 satisfies we need to find another value isn't it otherwise we can't claim insufficiency? How do we find the other number quickly... 3,23,43,83 and 103 don't satisfy the 1st equation
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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06 Oct 2013, 03:24
fozzzy wrote:
Bunuel wrote:
When the positive integer n is divided by 25, the remainder is 13. What is the value of n?

Given that $$n=25q+13$$, so n could be 13, 38, 63, 88, 113, ...

(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.

(2) When n is divided by 20, the remainder is 3 --> $$n=20p+3$$. n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.

(1)+(2) The only value of n which both statements is 63. Sufficient.

Out of the highlighted values only 63 satisfies we need to find another value isn't it otherwise we can't claim insufficiency? How do we find the other number quickly... 3,23,43,83 and 103 don't satisfy the 1st equation

Can you please elaborate what you mean? Thank you.
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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06 Oct 2013, 03:33
What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113

statement 2

n=20p + 3 => 3,23,43,63,83,103

For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?

don't we have to match the equations here?
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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06 Oct 2013, 03:47
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fozzzy wrote:
What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113

statement 2

n=20p + 3 => 3,23,43,63,83,103

For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?

don't we have to match the equations here?

Well, there cannot be only one value that satisfies both $$n=25q+13$$ (13, 38, 63, 88, 113, ...) and $$n=20p+3$$ (3, 23, 43, 63, 83, 103, ...).

Next, there is a way to derive general formula for $$n$$ (of a type $$n=mx+r$$, where $$x$$ is a divisor and $$r$$ is a remainder) based on above two statements:

Divisor $$x$$ would be the least common multiple of above two divisors 25 and 20, hence $$x=100$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=63$$.

Therefore general formula based on both statements is $$n=100m+63$$. Hence n can be 63, 163, 263, ...

Hope this helps.
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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06 Oct 2013, 03:58
Really neat trick... I'm gonna save this absolutely brilliant... Thanks Bunuel!
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Re: When the positive integer n is divided by 25, the remainder [#permalink]

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Re: When the positive integer n is divided by 25, the remainder   [#permalink] 13 Feb 2015, 11:53
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