How many positive integral values of N are there such that

Question

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

AnisMURR · Accepted Answer

\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.

hibobotamuss · Answer

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"

AnisMURR · Answer

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.

Mansoor50 · Answer

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

vishalkazone · Answer

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

Sent from my iPad using  GMAT Club Forum mobile app

Mansoor50 · Answer

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

AnisMURR · Answer

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.

ShankSouljaBoi · Answer

chetan2u  Any other approach ?

chetan2u · Answer

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   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.

Crytiocanalyst · Answer

N+4 +72 / N+4 = 1+ 72/N+4

Total number of divisors (3+1)*(2+1) = 12

As we only require possibilities greater than 4 since N is poditive and 0 is neither positive or negative

12-4 = 8

THerefore IMO C

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How many positive integral values of N are there such that

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