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N divided by 20 and 25 - GMATPrep

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Re: N divided by 20 and 25 - GMATPrep

by lion147 » Sun Jul 13, 2008 6:06 pm
AleksandrM wrote:Image
1) n<100, this means n has three possible values: 38, 63, and 88
not sufficient

2) Here we know n / 25 has the remainder 13, and n / 20 has the remainder 3.

So if...

n=38, n/20 = remainder 18
n=63, n/20 = remainder 3
n=88, n/20 = remainder 8
n=113, n/20 = remainder 13
n=138, n/20 = remainder 18
n=163, n/20 = remainder 3
etc...
not sufficient

1&2) Now you only have one value that satisfies the requirements, n<100, n/20=remainder 3 and n/25 = remainder 13; n must be 63.
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by AleksandrM » Mon Jul 14, 2008 8:43 am
lion,

While this approach certainly works, I am also very curious to see an algebraic approach.
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by gpranesh » Thu Jul 17, 2008 4:45 am
AleksandrM wrote:lion,

While this approach certainly works, I am also very curious to see an algebraic approach.
This is a kind of algebraic approach.

From initial data n / 25 has the remainder 13
-> n = 25*a + 13 , where a is positive integer (quotient)

From 1st stat, n < 100
-> 25*a+13 < 100
-> a < 87/25
-> a < 3.48 , Not Sufficient

From 2nd stat,
-> n = 20*a1 + 3
-> 25*a + 13 = 20*a1 + 3
-> 5*a = 4*a1 - 2

Hence, 4*a1-2 is multiple of 5, Since a,a1 are positive integers
a1 5*a
1 2
2 6
3 10 <-> a = 2
4 14
5 18
6 22
7 26
8 30 <-> a = 6

Not Sufficient, Since multiple solution
Combining stat 1 and stat 2, n<100
we get n = 25*2 + 13 = 63.
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