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.
_________________