Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Official Answer and Stats are available only to registered users. Register/Login.

_________________

Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690): http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

Last edited by Bunuel on 07 Jan 2013, 03:05, edited 1 time in total.

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

06 Jan 2013, 09:38

2

This post received KUDOS

1

This post was BOOKMARKED

2,3,5 are dinstinct PRIME factors...we'r asked for distinct factors... we get minimun value of y as 2*3*5=30...and 30 has 8 following distinct factors....1,2,3,5,6,10,15 and 30...hope the answer is clear now...:)

Posted from my mobile device _________________

Don't give up on yourself ever. Period. Beat it, no one wants to be defeated (My journey from 570 to 690): http://gmatclub.com/forum/beat-it-no-one-wants-to-be-defeated-journey-570-to-149968.html

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 20:50

2

This post received KUDOS

3

This post was BOOKMARKED

We need to find y with minimum possible value as the problem needs minimum distinct factors.

Prime factorization of \(60=2^2*3^1*5^1\)

As \(y^4\) is divisible by 60, it will include above prime factors of 60 (i.e. 2, 3 & 5) and we need to raise each prime factor to the power of 4 to get minimum \(y^4\) Minimum possible value of \(y^4 = (2^2*3^1*5^1) * (2^2*3^3*5^3) = (2^4*3^4*5^4)\) --This one must be divisible by 60.

Hence\(y = 2^1*3^1*5^1\)

Distinct factors of y = \((1+1)*(1+1)*(1+1) = 8\) --multiply (power of each prime factor +1)

Hence answer choice(C) is correct. _________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

06 Jan 2013, 11:41

1

This post received KUDOS

1

This post was BOOKMARKED

I feel that here it must be given that y is an integer. Anyways, an alternative approach is: To find the number of distinct factors of a number, first prime factorize it. In this case, since its given that \(y^4\) is a multiple of 60, hence \(y^4\) must contain 2*2*3*5. But here taking the fourth root will yield y in decimal form. Henceforth, to make y an integer, \(y^4\) must be atleast \(2^4 * 3^4 * 5^4\). Now since y is an integer and has 2,3 and 5 as its prime factors, so total number of prime factors will be 2*2*2=8. Since the number of prime factors is the product of the (power+1) of the individual prime factor. Here the individual powers are 1, 1 and 1. Hence the number of prime factors will be (1+1)*(1+1)*(1+1) or 8. Answer.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

06 Jan 2013, 11:43

1

This post received KUDOS

ConnectTheDots wrote:

daviesj wrote:

If \(y^4\) is divisible by 60, what is the minimum number of distinct factors that y must have? (A) 2 (B) 6 (C) 8 (D) 10 (E) 12

60= \(2^2*3^1*5^1\)

Number of distinct factors = (2+1)(1+1)(1+1) = 3*2*2 = 12 1,2,3,4 5,6,10,12, 15,20,30,60

Formula: If N = \(a^p*b^q*c^r...\), where a,b,c are prime numbers then Number of distinct factors = (p+1)(q+1)(r+1)

Why the answer is 8 ? What am I missing here ?

Here you are finding the distinct factors of \(y^4\) and not y. Rest of the method is correct. Moreover, I feel that it should be mentioned that y is an integer.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 21:48

1

This post received KUDOS

shanmugamgsn wrote:

Marcab wrote:

I feel that here it must be given that y is an integer. Anyways, an alternative approach is: To find the number of distinct factors of a number, first prime factorize it. In this case, since its given that \(y^4\) is a multiple of 60, hence \(y^4\) must contain 2*2*3*5. But here taking the fourth root will yield y in decimal form. Henceforth, to make y an integer, \(y^4\) must be atleast \(2^4 * 3^4 * 5^4\). Now since y is an integer and has 2,3 and 5 as its prime factors, so total number of prime factors will be 2*2*2=8. Since the number of prime factors is the product of the (power+1) of the individual prime factor. Here the individual powers are 1, 1 and 1. Hence the number of prime factors will be (1+1)*(1+1)*(1+1) or 8. Answer.

Marcab, Shouldn't this be \(2^8 * 3^4 * 5^4\) since in y there are \(2^2 * 3^1 * 5^1\) ? Please explain where im going wrong

See shan, We don't have to multiply the respective powers of each prime number by 4. We just have to multiply the powers with the smallest number so that together the product becomes the multiple of 4. Thats why I multiplied \(2^2\) with \(2^2\), \(3^1\) with \(3^3\) and \(5^1\) with \(5^3\). The resulting product became the multiple of 60 and when one takes fourth root, it become \(y=2*3*5\).

In the case \(2^8 * 3^4 * 5^4\), if we take the fourth root, the result will be \(2^2 * 3 *5\) and hence the number of prime factors will be \(3*2*2\) or 12. This is not the smallest. Hence incorrect.

If y^4 is divisible by 60, what is the minimum number of distinct factors that y must have?

(A) 2 (B) 6 (C) 8 (D) 10 (E) 12

We are given y^4/60 = integer. In other words:

y^4/(2^2 x 3^1 x 5^1) = integer

Since y must have at least one 2, one 3 and one 5 in order for y^4/60 = integer, the minimum value of y must be (2^1 x 3^1 x 5^1), or 30.

Now, to determine the number of distinct factors, we can use the following shortcut:

The total number of factors of a number can be obtained by multiplying the numbers resulting from adding 1 to the exponents in the prime factorization. Thus, the total number of factors of y is:

(1 + 1) x (1 + 1) x (1 + 1) = 2 x 2 x 2 = 8

Alternately, we could list all factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

Thus, y has 8 distinct factors.

Answer: C
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

If \(y^4\) is divisible by 60, what is the minimum number of distinct factors that y must have? (A) 2 (B) 6 (C) 8 (D) 10 (E) 12

60= \(2^2*3^1*5^1\)

Number of distinct factors = (2+1)(1+1)(1+1) = 3*2*2 = 12 1,2,3,4 5,6,10,12, 15,20,30,60

Formula: If N = \(a^p*b^q*c^r...\), where a,b,c are prime numbers then Number of distinct factors = (p+1)(q+1)(r+1)

Why the answer is 8 ? What am I missing here ?

Here you are finding the distinct factors of \(y^4\) and not y. Rest of the method is correct. Moreover, I feel that it should be mentioned that y is an integer.

That's correct. More precisely, it must be mentioned that y is a positive integer.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

08 Jan 2013, 18:59

I think I am understanding this correctly, but a little confused.

Maybe if we change things up a little bit, I can see how this works: If instead of 60, Y was 210, what would the answer be? How would you arrive to the conclusion?

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

08 Jan 2013, 19:24

hitman5532 wrote:

I think I am understanding this correctly, but a little confused.

Maybe if we change things up a little bit, I can see how this works: If instead of 60, Y was 210, what would the answer be? How would you arrive to the conclusion?

If Y were 210, then first step would have been finding the prime factors. 210=2*5*3*7

The total number of disntict factors would be 2*2*2*2=16.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 03:12

1

This post was BOOKMARKED

Marcab wrote:

I feel that here it must be given that y is an integer. Anyways, an alternative approach is: To find the number of distinct factors of a number, first prime factorize it. In this case, since its given that \(y^4\) is a multiple of 60, hence \(y^4\) must contain 2*2*3*5. But here taking the fourth root will yield y in decimal form. Henceforth, to make y an integer, \(y^4\) must be atleast \(2^4 * 3^4 * 5^4\). Now since y is an integer and has 2,3 and 5 as its prime factors, so total number of prime factors will be 2*2*2=8.Since the number of prime factors is the product of the (power+1) of the individual prime factor. Here the individual powers are 1, 1 and 1. Hence the number of prime factors will be (1+1)*(1+1)*(1+1) or 8. Answer.

Hey, Marcab,I still dont get the quoted part in ur statement...

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 03:50

bhavinshah5685 wrote:

Marcab wrote:

I feel that here it must be given that y is an integer. Anyways, an alternative approach is: To find the number of distinct factors of a number, first prime factorize it. In this case, since its given that \(y^4\) is a multiple of 60, hence \(y^4\) must contain 2*2*3*5. But here taking the fourth root will yield y in decimal form. Henceforth, to make y an integer, \(y^4\) must be atleast \(2^4 * 3^4 * 5^4\). Now since y is an integer and has 2,3 and 5 as its prime factors, so total number of prime factors will be 2*2*2=8.Since the number of prime factors is the product of the (power+1) of the individual prime factor. Here the individual powers are 1, 1 and 1. Hence the number of prime factors will be (1+1)*(1+1)*(1+1) or 8. Answer.

Hey, Marcab,I still dont get the quoted part in ur statement...

I got answer 12.

Hii Bhavin. Its given that \(y^4\) is a multiple of 60. So \(y^4\) must be atleast 60 or \(2^2 * 3 * 4\). Taking the fourth root will result: \(2^{1/2} * 3^{1/4} * 5^{1/4}\). Since neither of \(2^{1/2}\) ,\(3^{1/4}\) and \(5^{1/4}\) is an integer, therefore fourth root will yield decimal number. To get y as an integer, the powers of 2,3 and 5 must be a multiple of 4, so that the fourth root yields an integer. hope that helps.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 04:03

12 can't be the answer. Correct answer is 8. First make prime factorization of an integer n=\(a^p * b^q * c^r\), where a, b, and c are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of will be expressed by the formula \((p+1)*(q+1)*(r+1)\). NOTE: this will include 1 and n itself.

Concentration: Entrepreneurship, International Business

GMAT 1: 440 Q33 V13

GPA: 3

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

09 Jan 2013, 20:22

Marcab wrote:

I feel that here it must be given that y is an integer. Anyways, an alternative approach is: To find the number of distinct factors of a number, first prime factorize it. In this case, since its given that \(y^4\) is a multiple of 60, hence \(y^4\) must contain 2*2*3*5. But here taking the fourth root will yield y in decimal form. Henceforth, to make y an integer, \(y^4\) must be atleast \(2^4 * 3^4 * 5^4\). Now since y is an integer and has 2,3 and 5 as its prime factors, so total number of prime factors will be 2*2*2=8. Since the number of prime factors is the product of the (power+1) of the individual prime factor. Here the individual powers are 1, 1 and 1. Hence the number of prime factors will be (1+1)*(1+1)*(1+1) or 8. Answer.

Marcab, Shouldn't this be \(2^8 * 3^4 * 5^4\) since in y there are \(2^2 * 3^1 * 5^1\) ? Please explain where im going wrong
_________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

12 Nov 2014, 10:27

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

10 Feb 2017, 02:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: If y^4 is divisible by 60, what is the minimum number of dis [#permalink]

Show Tags

10 Feb 2017, 06:06

I got 6 as ans Since factor of 60 2^2*3*5 and since minimum factor of y is asked so it much contain one nos of 2,3,5 so the min no of distinct factors are 6.