A positive integer x has 60 divisors and 7x has 80 divisors. What is t

I am still unable to understand this explanation..
Total number of factors of x = 60 = 2^2 * 3 * 5(Its fine till here) = (3+1)*(2+1)*(4+1)(why this??)
Total number of factors of 7x = 80 = 2^2 * 4 * 5(Its fine till here) = (3+1)*(3+1)*(4+1)(why this??)

And this part leaves me

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.

Could you please elaborate on this?How can a 3 become a 4?and why?and how does 7 come into the picture?

TIA!

This tells you that if N is of the form a^p*b^q*c^r..., the Total Number of Factors of a number N = (p+1)(q+1)(r+1)... (note that this is the number of factors of N, not N itself)

Here you are given that total number of factors on x is 60. Remember, x is not 60. The total number of factors of x is 60.
You need to write 60 in the form (p+1)(q+1)(r+1)...
One way of doing that is \(60 = 4*3*5 = (3+1)*(2+1)*(4+1)\) Note that 4 = 3+1, 3 = 2+1, 5 = 4+1. So we have done nothing except changed the form.

When will we write that total number of factors of x are (3+1)*(2+1)*(4+1)? This happens when x is of the form \(a^3 * b^2 * c^4\)

hi chetan2u,
super bro.. how could simplify it so easily ..

Great Karishma. I also read your post and it was quite helpful.

One question I have refers to the difference of x and 7x. What do we mean with "7x has only one 7 more than x" and with "Only the power of 7 increases by 1"?

I do understand the difference between (3+1)*(2+1)*(4+1) and (3+1)*(3+1)*(4+1), and I can see that the 2 became a 3. But I don't know how this relates to 7...

Ask yourself, why does 2 become 3?

Say \(x = 2^3 * 7^2 * 11^4\)
Total number of factors of \(x = (3+1)*(2+1)*(4+1) = 60\)

What will be 7x?

\(7x = 2^3 * 7^3 * 11^4\)
When you prime factorize 7x, you will get another 7.
Total number of factors of \(7x = (3+1)*(3+1)*(4+1) = 80\)

So, the point it that you can get whatever number using this number of factors for x: (3+1)*(2+1)*(4+1).
Then, you see than when x becomes 7x, the number of factors of one of the bases increases: (3+1)*(3+1)*(4+1).

Since the only thing that changed is that 7x has an additional prime factor (7), we conclude that this 7 belonged to (2+1) that became (3+1). Then we know that it was 2 before and this is how we get to our answer.

How is n related to x? I didn't understand this question at all..

Here's another method to do this. I have used algebraic method here..
First of all..notice that a power of 7 has to be a part of n already for the given condition to be true, because if that is not the case..then a completely new 7 will make the number of factors as 120(exactly double).
The number of factors of n can be written as
(a+1)(b+1)(c+1)...(s+1)(t+1) = 60...(t+1) is for powers of 7 ------ eq(1)

Let (a+1)(b+1)...(s+1) = k

According to the next condition..
(a+1)(b+1)....(s+1)(t+2) = 80

Or

k(t+2) = 80
=> k(t+1+1) = 80
=> k(t+1) + k = 80
Using the given value
=> 60 + k = 80
=> k = 20
Using this value in eq(1)

(t+1) = 3
=> t = 2

And Voila!
Answer(C)

Note that n is a typo. The question is "What is the greatest integer y such that 7^y divides x?

As I understand there should be x instead of n.
If x has 60 divisors we can represent it as follows:

1*60; 2*30; 3*20; 4*15 ; 5*12 ; 6*10 ; 2*2*15 ; 2*3*10 ; 2*2*3*5
\(p^{59}\) ; \(p*q^{29}\) ; \(p^2*q^{19}\) ; \(p^3*q^{14}\) ; \(p^4*q^{11}\) ... (etc. - power of prime is 1 less than its representation in factor grouping)

Now if 7 is not present in initial number x, then in 7*x we’ll have 120 as total number of factors (we are adding 2 additional choices for additional prime and due to multiplication rule the whole result will be doubled).

So the only possibility is that 7 is already a prime factor of x. In this case, we just need to increase its power by 1. The obvious choice we can spot at first glance is 3*20. In order to get 80 we need 4*20. So the initial power of 7 in x was one less than 3.

Answer 2 (C)

Kind of amateurish by I hope understandable

I solved it like this:

First thing to note: if x does not have 7 as one of the prime factors, then 7x will have double the factors i.e. 120

so x already has 7 as one of the prime factors.
say prime factorization of \(x = (p + 1)(q+1)(r+1).... = 60\) (total factors)
Let p be the power of 7 in x

then prime factorization of \(7x = (p+1+1)(q+1)(r+1).... = 80\)
substituting \((q+1)(r+1)....\) as = \(60/(p+1)\)
=> \((p+2) * 60/(p+1) = 80\)
=> \(3p + 6 = 4p + 4\) => \(p = 2\)
so x has \(7^2\) has the factor.

So maximum y such that \(7^y\) will divide x is 2 => (C)

A positive integer x has 60 divisors and 7x has 80 divisors. What is the greatest integer y such that \(7^y \)divides n

60= \(2^2\)*3*5 = (2+1)*(3+1)*(4+1)
80= \(2^4\)*5 = (3+1)*(3+1)*(4+1)

if 80 divisors for 7x, it means, only one power increased by 1.
So, 7^2.

I think C.