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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.

Re: A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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10 Feb 2015, 10:51

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?

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?

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 \(a^3 * b^2 * c^4\)
_________________

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
_________________

Re: A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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11 Feb 2015, 20:42

chetan2u wrote:

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

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

Re: A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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02 Mar 2015, 13:10

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)

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...

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 \(7x = (3+1)*(3+1)*(4+1) = 80\)
_________________

Re: A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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03 Mar 2015, 00:52

VeritasPrepKarishma wrote:

pacifist85 wrote:

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.

A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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31 Oct 2016, 03:53

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

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)

A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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31 Oct 2016, 09:06

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.

A positive integer x has 60 divisors and 7x has 80 divisors. What is t [#permalink]

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01 Jan 2018, 20:15

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)