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Last edited by Bunuel on 07 Jan 2013, 02:05, edited 1 time in total.

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

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06 Jan 2013, 08:38

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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...:)

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

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06 Jan 2013, 10:41

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

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06 Jan 2013, 10:43

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

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08 Jan 2013, 17:59

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]

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08 Jan 2013, 18:24

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

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09 Jan 2013, 02:12

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]

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09 Jan 2013, 02:50

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

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09 Jan 2013, 03:03

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.

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

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09 Jan 2013, 19:22

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

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09 Jan 2013, 19:50

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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]

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09 Jan 2013, 20:48

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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.

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