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 themour 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.

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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