If y^4 is divisible by 60, what is the minimum number of dis

I get 3, since 60^4 is divisable through 60 and 60 has only 3 distinct factors which are 2, 3 and 5...

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 ?

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.

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.

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.

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.

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.

For more on number theory, do visit:
math-number-theory-88376.html

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.

Hope that helps.

minimum value y can be is 2*3*5. all raised to the power of 4 is divisible by 60.
thus, to find the number of factors that y can have:
2*2*2 = 8.
C.

bumped into this one again. did the same way - under 50 seconds.

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.



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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

Hi all,

I read through all responses but I'm still having some trouble understanding this. Going by the formula... Number of distinct factors = (p+1)(q+1)(r+1)... I have 2^2, 3^1, 5^1 - where p =2, q = 1, and r = 1. My initial reaction is to plug in and get 12. I understand that we need y, not y^4, but why does that change things?

I guess my question is, why exactly are we making each power 1 (changing 2^2 to 2^1) in this case? When would we do this vs when would we not do this? I need to understand what makes this problem unique so that on test day, I don't get a question like this wrong.

If it was y is divisible by 60 (instead of y^4), then would the answer be 12? Also, if it was y^2 or y^3 instead of y^4, would that change things?

Thank you very much to whoever can set me straight on this!

Best,
Max

in this question, the assumption is that y is an integer. Then, the question can be paraphrased into the following idea: if y is a minimum integer that y^4 can be divisible by 60 , how many distinct factors y has?

3 distinct prime factors of the minimum y are 2,3,5 => distinct factors will be 8