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 Post subject: When positive integer n is divided by 25, the remainder....
 Post Posted: Wed Apr 23, 2008 6:37 am 
When 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

C*

How should I think about remainders? I find these types of problems confusing.


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 Post subject:
 Post Posted: Thu Apr 24, 2008 2:03 am 
When 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



N/25=x+13 --> N= 25x+13 ---(Given)

What is N?

ADBCE Grid:

A) N<100, which means N can have any value from 1 to 99. Insuff
B) N/20= X+3 --> N= 20X+3 Where N Can have any number of values. INsuff

Using A and B,
N<100 and N = 20X+3

Suff to Find the value of N.

Hence C

Thanks


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 Post subject:
 Post Posted: Thu Apr 24, 2008 4:55 pm 
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ManhattanGMAT Staff


Posts: 384
Quote:
When 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


Yes, Sudhan, but it's not clear from your solution why (1&2) is sufficient. Here's a numerical approach.

Rephrased question:
The possible values of N are 13, 38, 63, 88, 113, 138, 163, 188, 213... what is n?

(1) Not enough information, as there are many values less than 100 in our list.
(2) So, N could be 3, 23, 43, 63, 83, 103, 123, 143, 163, 183, ... Again, we have at least two cases: 63 and 163. Insufficient.
(1&2) This narrows it down to exactly one value: 63. Sufficient.

Another way to deal with it is to express N as functions.

Rephrased question:
N = 25x + 13. What is N? (Here, the variable x can take on integer values. For each x, you'll get a corresponding value of N.)

(1) N < 100 is not sufficient. If x is 1, for example, N is 38 and if x is 3, N is 88.
(2) We are told that N can be expressed as 20y + 3 (where y is an integer). So we can attempt to solve the system:
N = 20y + 3
N = 25x + 13
Can't do it! Only two equations and three variables. Insufficient.

(1&2) Now let's go back to that system from (2). It yields:
20y + 3 = 25x + 13
20y = 25x + 10
So now it's a matter of finding values of x and y that make the equation true. If x = 1, the right side is 35, and that's not a multiple of 20 (as required by the left side). If x = 2, the right side is 60, and that *is* a multiple of 20. So x = 2 and y = 3 is a viable solution. That corresponds to a value of N = 63. Keep going, though. If x = 3, the right side is 85 -- no good, since 88 is not a multiple of 20. If you keep going with this exercise, the next value of x that works is x = 6, since y would equal 8. BUT, x = 6 corresponds to a value of N = 163, which is bigger than 100. Sufficient.

I think you'd agree that for these problems, a numerical solution is probably most efficient. But either way gets you there.

Rey


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 Post subject:
 Post Posted: Thu May 01, 2008 12:27 am 
Rey,

In above solution, once you find first value of N which satisfy both the eqn, You can add LCM of 20 and 25 to first value. So other values of N will be 63, 163, 263 etc
I find this method much faster but want to check whether this can be applied to any remainder equation.


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 Post subject:
 Post Posted: Wed May 07, 2008 5:57 am 
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ManhattanGMAT Staff


Posts: 13427
Path wrote:
Rey,

In above solution, once you find first value of N which satisfy both the eqn, You can add LCM of 20 and 25 to first value. So other values of N will be 63, 163, 263 etc
I find this method much faster but want to check whether this can be applied to any remainder equation.


that will work, because you have to add a multiple of N for the remainder upon division by N to stay the same.

since you want both remainders to stay the same, you must add a number that's a multiple of both numbers - so your lcm approach works.

on the other hand, it also shouldn't take you more than a few seconds to generate the lists that rey (rfernandez) has shown above.
if you actually do find this method "much faster", that's a sign that you should get better at generating that kind of list.


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 Post subject:
 Post Posted: Wed May 14, 2008 5:31 pm 
Here it is useful to see that both 20 and 25 are divisors of 100

Multiples of 25 have the tens and units digits 00,25,50,75
So the questions tells us that n ends in either 13,38,63 or 78

Similarly, (2) tells us that n ends in 03, 23, 43,63 or 83

Kevin Armstrong,
Madrid, Spain


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 Post subject:
 Post Posted: Fri Jun 06, 2008 4:27 am 
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ManhattanGMAT Staff


Posts: 384
Good work everyone.


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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Wed Nov 30, 2011 5:12 am 
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Forum Guests


Posts: 618
Rey,

In above solution, once you find first value of N which satisfy both the eqn, You can add LCM of 20 and 25 to first value. So other values of N will be 63, 163, 263 etc
I find this method much faster but want to check whether this can be applied to any remainder equation.

yes, the LCM above can be used for many remainder questions on gmatprep. but inhere the LCM can not be used. only pick the number.

I am not sure, pls, give an example in which the LCM can be used

for exmple, if n is divided by 6 the remainder is 2, if n is divided by 10 the remainder is 3. so n=LCM (6,10)+ min number which satisfy the 2 condition above , and is found by picking number.

I do not know how to do more, pls, help


n=30k+


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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Fri Dec 09, 2011 4:38 am 
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ManhattanGMAT Staff


Posts: 13427
thanghnvn, i'm sorry, but i can't discern the actual question you are trying to ask.

could you please ask a question (in, say, 1-2 lines of text)?
thanks.

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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Sat Jan 26, 2013 2:57 am 
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Posts: 175
Location: Bangalore
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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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Sat Jan 26, 2013 7:04 am 
Offline
ManhattanGMAT Staff


Posts: 13427
sachin.w wrote:
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


fine, but, also make sure you can generate lists of the sort given by rey (rfernandez) above, without having to think too much. there are lots of remainder problems on which you're best served by just pounding out lists of numbers, so it's an indispensable skill (much more important on this exam, in fact, than any algebraic approach to remainders).

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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Sat Jan 26, 2013 10:55 pm 
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Posts: 175
Location: Bangalore
sure thing. thanks a lot, Ron :)


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 Post subject: Re: When positive integer n is divided by 25, the remainder....
 Post Posted: Sat Jan 26, 2013 11:04 pm 
Offline
ManhattanGMAT Staff


Posts: 2648
sachin.w wrote:
sure thing. thanks a lot, Ron :)


:-)

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Jamie Nelson
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