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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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03 Nov 2018, 09:57
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95% (hard)

Question Stats:

33% (02:19) correct 67% (02:36) wrong based on 118 sessions

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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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04 Nov 2018, 02:28
4
6
$$\frac{N+76}{N+4}= \frac{N+4+72}{N+4}=1+\frac{72}{N+4}$$

$$1$$ 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 $$12-4=8$$

Don't hesitate to ask if something is not clear.
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Re: How many positive integral values of N are there such that  [#permalink]

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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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05 Nov 2018, 04:30
1
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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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 $$12-4=8$$

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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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 $$12-4=8$$

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

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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 $$12-4=8$$

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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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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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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11 Nov 2018, 09:51
2
1
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

(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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