VeritasPrepKarishma wrote:
manpreetsingh86 wrote:
A positive integer x has 60 divisors and 7x has 80 divisors. What is the greatest integer y such that \(7^y\) divides n?
a) 0
b) 1
c) 2
d) 3
e) 4
Total number of factors of x = 60 = (p+1)*(q+1)*(r+1)... = 2^2 * 3 * 5
Now note that 7x has only one 7 more than x. The number of all other prime factors stays the same.
Total number of factors of 7x = 80 = (p+2)*(q+1)*(r+1)... = 2^4 * 5
Here, the 3 of previous expression has disappeared so it must have converted to 4. Does it make sense? Let's see:
Total number of factors of x = 60 = 2^2 * 3 * 5 = (3+1)*(2+1)*(4+1)
Total number of factors of 7x = 80 = 2^2 * 4 * 5 = (3+1)*(3+1)*(4+1)
Perfect!
The powers of other factors stay the same. Only the power of 7 increases by 1. So initially, in x, the power of 7 must have been 2. Hence the maximum value of y must be 2.
Answer (C)
hi,
you have given a good way of solving it .. kudos for that
i can think of one straight way to do it..
let the number(x..n..?) be a^k*b^l...7^t..
here we are just interested in value of 't'.
two points now..
1) n=a^k*b^l...7^t.. so number of factors=(k+1)(l+1)..(t+1)=60..
lets take all other values as z ie z=(k+1)(l+1).....
so z(t+1)=60....(1)
2)7n=a^k*b^l...7^(1+t).. so number of factors=(k+1)(l+1)..(t+2)=80..
as we take all other values as z ie z=(k+1)(l+1).....
so z(t+2)=80....(2)
from eq (1)and(2)...
80(t+1)=60(t+2)... t=2..
so ans is 2.. C
super bro.. how could simplify it so easily ..