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If m and n are positive integers such that m is a factor of

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If m and n are positive integers such that m is a factor of [#permalink] New post 04 Oct 2011, 19:59
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If m and n are positive integers such that m is a factor of n, how many positive multiples of m are less than or equal to 2n ?

A. 2m/n + 1
B. 2n/m + 1
C. 2n/(m+1)
D. 2m/n
E. 2n/m

what is the best method to approach problems like these ? - Algebraic methods or substitution method ? Any advice ?
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Re: How many factors are there ? [#permalink] New post 04 Oct 2011, 20:10
kulki wrote:
If m and n are positive integers such that m is a factor of n, how many positive multiples of m are less than or equal to 2n ?
a. 2m/n + 1
b. 2n/m + 1
c. 2n/(m+1)
d. 2m/n
e. 2n/m

what is the best method to approach problems like these ? - Algebraic methods or substitution method ? Any advice ?


Lets say N=10, M=5
2N=20. so the answer should be 4 (20/5)
lets try to plug in the answers:
A-not an integer
B-not an integer
C-not an integer
D-1 (not the answer)
E-4 - the answer. (the only one).

I would choose E.

Method 2
N=M*A (A is an integer)
So - A=N/M
therefore in 2N A will be 2N/M

Again - Answer is E.

Hope it helps.

*** personally - I love the mathematical approach. Easier for me, however, many people like the plug in numbers approach. It is important to master both methods so if one doesnt work, you can try the other.
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Re: How many factors are there ? [#permalink] New post 06 Oct 2011, 03:25
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kulki wrote:
If m and n are positive integers such that m is a factor of n, how many positive multiples of m are less than or equal to 2n ?
a. 2m/n + 1
b. 2n/m + 1
c. 2n/(m+1)
d. 2m/n
e. 2n/m

what is the best method to approach problems like these ? - Algebraic methods or substitution method ? Any advice ?


I don't think there is anything called 'the best approach'. Each person uses what works best for her/him. I jump to plugging in numbers whenever I see variables in the questions and variables in the options. I know that within 1 or 2 iterations, I should get my answer. I also like to choose the simplest possible numbers. e.g. here I would put m = 1 and n = 1 (both positive integers and m is a factor of n). But the moment I put them equal, I see that options d and e will be the same. So instead, I put m = 1 and n = 2. 2n = 4.
Number of positive multiples of 1 that are less than or equal to 4 = 4 (1, 2, 3, 4)
Only e gives me 4 when I put n = 2 and m = 1
Answer (E).

Mind you, it is not a good idea to plug in numbers when you have 3-4 variables since you might just get lost in them.
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Re: How many factors are there ? [#permalink] New post 06 Oct 2011, 08:33
Karishma, you are the greatest. +1
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Re: How many factors are there ? [#permalink] New post 06 Oct 2011, 09:05
thank you for this!
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Re: How many factors are there ? [#permalink] New post 07 Oct 2011, 09:34
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VeritasPrepKarishma wrote:
kulki wrote:
If m and n are positive integers such that m is a factor of n, how many positive multiples of m are less than or equal to 2n ?
a. 2m/n + 1
b. 2n/m + 1
c. 2n/(m+1)
d. 2m/n
e. 2n/m

what is the best method to approach problems like these ? - Algebraic methods or substitution method ? Any advice ?


I don't think there is anything called 'the best approach'. Each person uses what works best for her/him. I jump to plugging in numbers whenever I see variables in the questions and variables in the options. I know that within 1 or 2 iterations, I should get my answer. I also like to choose the simplest possible numbers. e.g. here I would put m = 1 and n = 1 (both positive integers and m is a factor of n). But the moment I put them equal, I see that options d and e will be the same. So instead, I put m = 1 and n = 2. 2n = 4.
Number of positive multiples of 1 that are less than or equal to 4 = 4 (1, 2, 3, 4)
Only e gives me 4 when I put n = 2 and m = 1
Answer (E).

Mind you, it is not a good idea to plug in numbers when you have 3-4 variables since you might just get lost in them.


In case of 3-4 variables good to use the table to track.
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Re: How many factors are there ? [#permalink] New post 08 Oct 2011, 23:41
I think picking number is the best for this sort of problems.
i Picked m=3, n=12
Only E worked.
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Re: How many factors are there ? [#permalink] New post 14 Oct 2011, 01:15
@karishma..

Same here i plugged in numbers too..m=2 and n=6..

IMO E..
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Re: How many factors are there ? [#permalink] New post 18 Oct 2011, 20:24
Plugging numbers seems like the best approach here.

1. Put n = 25 and m = 5
2. Put n = 4 and m = 2.

In both the scenarios only E stands out.
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Re: How many factors are there ? [#permalink] New post 03 Jun 2012, 08:18
kulki wrote:
If m and n are positive integers such that m is a factor of n, how many positive multiples of m are less than or equal to 2n ?
a. 2m/n + 1
b. 2n/m + 1
c. 2n/(m+1)
d. 2m/n
e. 2n/m

what is the best method to approach problems like these ? - Algebraic methods or substitution method ? Any advice ?


I think we can save some time here. It is a simple question asking the number of factors but in a twisted manner. For example, if I may ask how many positive multiples of 3 are less than or equal to 12 (6*2), given that 6 is multiple of 3. Its straight 12/3.
Re: How many factors are there ?   [#permalink] 03 Jun 2012, 08:18
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