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How many positive integral values of N are there such that

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How many positive integral values of N are there such that  [#permalink]

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New post 03 Nov 2018, 09:57
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How many positive integral values of N are there such that \(\frac{N+76}{N+4}\) is an integer?

a) 5
b) 7
c) 8
d) 9
e) 10

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Re: How many positive integral values of N are there such that  [#permalink]

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New post 04 Nov 2018, 02:28
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\(\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}\)

\(1\) is an integer, so let's see for how many values of N, \(\frac{72}{N+4}\) is an integer.

\(72 = 2^3*3^2\)

Thus the number of factors (numbers that can divide \(72\)) is equal to \((3+1)*(2+1)=12\)

So \(72\) has \(12\) factors, but \(N+4\) is greater than \(4\) so we have to remove the factors that are less or equal to \(4\).

So we have to remove, \(1\), \(2\), \(3\) and \(4\) (those are the factors of \(72\) that are less or equal than \(4\))

So \(12-4=8\)

Answer C


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Re: How many positive integral values of N are there such that  [#permalink]

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New post 05 Nov 2018, 03:16
AnisMURR How did you get "Thus the number of factors (numbers that can divide 7272) is equal to (3+1)∗(2+1)=12(3+1)∗(2+1)=12"
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 05 Nov 2018, 04:30
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hibobotamuss

\(72=2^3*3^2\)

You take all the powers of the prime factors you add 1, and then you multiply them. It is a formula.

lets say \(X=2^3*3^2*5^5*7^4\)

The number of factors is \(4*3*6*5\)

Don't hesitate if it is not clear, you can also visit my youtube channel for more exercices.
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 05 Nov 2018, 18:28
AnisMURR wrote:
\(\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}\)

\(1\) is an integer, so let's see for how many values of N, \(\frac{72}{N+4}\) is an integer.

\(72 = 2^3*3^2\)

Thus the number of factors (numbers that can divide \(72\)) is equal to \((3+1)*(2+1)=12\)

So \(72\) has \(12\) factors, but \(N+4\) is greater than \(4\) so we have to remove the factors that are less or equal to \(4\).

So we have to remove, \(1\), \(2\), \(3\) and \(4\) (those are the factors of \(72\) that are less or equal than \(4\))

So \(12-4=8\)

Answer C


Don't hesitate to ask if something is not clear.


when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 05 Nov 2018, 18:55
Mansoor50 wrote:
AnisMURR wrote:
\(\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}\)

\(1\) is an integer, so let's see for how many values of N, \(\frac{72}{N+4}\) is an integer.

\(72 = 2^3*3^2\)

Thus the number of factors (numbers that can divide \(72\)) is equal to \((3+1)*(2+1)=12\)

So \(72\) has \(12\) factors, but \(N+4\) is greater than \(4\) so we have to remove the factors that are less or equal to \(4\).

So we have to remove, \(1\), \(2\), \(3\) and \(4\) (those are the factors of \(72\) that are less or equal than \(4\))

So \(12-4=8\)

Answer C


Don't hesitate to ask if something is not clear.


when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards

As per question Only positive value of N to be considered.


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Re: How many positive integral values of N are there such that  [#permalink]

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New post 06 Nov 2018, 04:24
vishalkazone wrote:
Mansoor50 wrote:
AnisMURR wrote:
\(\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}\)

\(1\) is an integer, so let's see for how many values of N, \(\frac{72}{N+4}\) is an integer.

\(72 = 2^3*3^2\)

Thus the number of factors (numbers that can divide \(72\)) is equal to \((3+1)*(2+1)=12\)

So \(72\) has \(12\) factors, but \(N+4\) is greater than \(4\) so we have to remove the factors that are less or equal to \(4\).

So we have to remove, \(1\), \(2\), \(3\) and \(4\) (those are the factors of \(72\) that are less or equal than \(4\))

So \(12-4=8\)

Answer C


Don't hesitate to ask if something is not clear.


when you say N+4 is greater than 4...you are excluding negative values of N.

Can you explain why I cant have N<0?

regards

As per question Only positive value of N to be considered.


Sent from my iPad using GMAT Club Forum mobile app


Thanks!!!!

note to self: Mansoor...read the question CAREFULLY!!!!

:)
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 06 Nov 2018, 04:53
vishalkazone
Mansoor50

"How many positive integral values of N " N should be positive

Don't hesitate if it is not clear, you can also visit my youtube channel for more exercices.
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 11 Nov 2018, 06:07
chetan2u Any other approach ?
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Re: How many positive integral values of N are there such that  [#permalink]

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New post 11 Nov 2018, 09:51
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Afc0892 wrote:
How many positive integral values of N are there such that \(\frac{N+76}{N+4}\) is an integer?

a) 5
b) 7
c) 8
d) 9
e) 10


Ok ShankSouljaBoi,

Two ways,
(I) I will go with this, as also discussed above and would recommend to anyone who is aware of the method..
\(\frac{N+76}{N+4}= \frac{N+4}{N+4} + \frac{72}{N+4}\)
Now 72 should be a multiple of N+4....

So, our next step should be to find the factors of 72.
\(72=2^3*3^2\).... Number of factors = (3+1)(2+1)=4*3=12
so N+4 can be equal to any of these 12 factors.

As, N is positive so N+4 will have the least value of 1+4 or 5, when N is 1.
Keeping in mind the constraint, all factors below 5 are not possible.
WHAT are the factors below 5 - they are 1, 2, 3, 4... so 4 of them
our answer 12-4=8

(II) If you are stuck..
any of such questions should be made equal to a variable.. so \(\frac{N+76}{N+4}=a.......N+76=aN+4a..........N(a-1)=76-4a.......N = \frac{76-4a}{a-1}\)
since N is integer, [\(fraction]76-4a/a-1[/fraction]\) must be an integer.
so FIRST condition - \frac{numerator}{geq {denominator}} thus \(76-4a/geq{a-1}........a<\frac{77}{5}......a<15.4\), thus greatest value of a is 15
so you have to just substitute a from 1 to 15 and see which all values give us an integer.

for example a = 1 not possible...
a=2, 76-4a=76-8=68 and a-1=2-1=1... 68 divisible by 1 ..yes
a=3, 76-4a=76-12=64 and a-1=3-1=2... 64 divisible by 2 ..yes
a=4, 76-4a=76-16=60 and a-1=4-1=3... 60 divisible by 3 ..yes
a=5, 76-4a=76-20=56 and a-1=5-1=4... 56 divisible by 4 ..yes
a=6, 76-4a=76-24=52 and a-1=6-1=5... 52 is not divisible by 3 ..
a=7, 76-4a=76-28=48 and a-1=7-1=6... 48 divisible by 3 ..yes
and so on..
Although it will not take too much time but let us see if we can shorten it..
when a is ODD, both are even so 3, 5, 7, 9, 13 will give you an integer.
they leave the same remainder when divided by 3 ... 76-4a or 75+1-3a-a will leave a remainder of 1-a as 75-3a will be divisible by 3 and the denominator is also a-1..
so when a is 4, 4+3, 7+3, 10+3 so 4,7,10,13
so 2, 3, 4, 5, 7, 9, 10, 13.. thus 8 of them ..
you can try for values of 6 or 11, but they both will not be divisible by 3 together.
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Re: How many positive integral values of N are there such that   [#permalink] 11 Nov 2018, 09:51
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