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If p^3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

2 3 6 8 10

Prime factorize \(80 = 2^4 * 5\) If \(p^3\) has at least four 2s and a 5, it must have at least six 2s and three 5s (Every prime factor of \(p^3\) must have a power which is a multiple of 3). So p must have at least two 2s and a 5 as factors.

Minimum value of \(p = 2^2 * 5\) This gives us \((2+1)*(1+1) = 6\) distinct factors (at least)
_________________

If p3 is divisible by 80, then the positive integer p must have at least how many distinct factors? (A) 2 (B) 3 (C) 6 (D) 8 (E) 10

Please someone explain this question with solution .Thanks

hi, the method to find distinct factors is.. step 1.. break down the integer in its basic form with prime numbers.. 80=2^4*5... step 2.. formula is\(a^x*b^y... (x+1)(y+1)\)... so here the answer will be (4+1)(1+1)=5*2=10 ans E.. hope it helped

but it seems you mean p3 as \(p^3\)... so p^3 will have atleast\(2^4*5\)as its factor, and therefore, p will have atleast \(2^2*5\) as factors.. ans 3*2=6 ans C
_________________

If p3 is divisible by 80, then the positive integer p must have at least how many distinct factors? (A) 2 (B) 3 (C) 6 (D) 8 (E) 10

Please someone explain this question with solution .Thanks

Since p is an Integer, therefore p^3 must a perfect cube

Perfect Cube is a number that has all the powers of its Prime factors a multiple of 3 when the Number is written in Prime factorized form

But \(p^3 = 80x = 2^4*5*x\)

i.e. The value of \(x\) must be a smallest number which can make p^3 a Perfect cube and keep the number smallest for Minimum number of factors of p

i.e. \(x_{min} = 2^2*5^2\)

such that \((p^3)_{min} = 2^4*5*2^2*5^2 = 2^6*5^3\)

i.e. \(p_{min} = 2^2*5\)

Number of Factors of \(2^2*5 = (2+1)*(1+1) = 3*2 = 6\)

Answer: Option C _________________

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Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink]

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30 May 2016, 08:05

VeritasPrepKarishma wrote:

ggarr wrote:

If p^3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

2 3 6 8 10

Prime factorize \(80 = 2^4 * 5\) If \(p^3\) has at least four 2s and a 5, it must have at least six 2s and three 5s (Every prime factor of \(p^3\) must have a power which is a multiple of 3). So p must have at least two 2s and a 5 as factors.

Minimum value of \(p = 2^2 * 5\) This gives us \((2+1)*(1+1) = 6\) distinct factors (at least)

I fail to understand what you mean by "every prime factor of p must have a power which is a multiple of 3". My guess is that as there are 3 p's, they must all have the same factors with powers and hence 2^4 and 5, have been considered as 2^6 and 5^3. so it can be evenly divided between 3 p's and their total of 8000 is divisible by P. Could you shared some light on the same.

Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink]

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30 May 2016, 08:19

Bunuel wrote:

Tough and Tricky questions: Factors.

If p^3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

(A) 2 (B) 3 (C) 6 (D) 8 (E) 10

80 = \(2^4\) x \(5^1\)

Since p^3 is divisible by \(2^4\) x \(5^1\) the least value of p will be \(2^6\) x \(5^3\) ; where \(p\) = \(5^1\) x \(2^2\) So, p must have (1+1) ( 2 + 1 ) => 6 factors, answer will be (C) _________________

Thanks and Regards

Abhishek....

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Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink]

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27 Dec 2016, 04:31

1

This post received KUDOS

Let's start by breaking 80 down into its prime factorization: 80 = 2 × 2 × 2 × 2 × 5. If p^3 is divisible by 80, p^3 must have 2, 2, 2, 2, and 5 in its prime factorization. Since p^3 is actually p × p × p, we can conclude that the prime factorization of p × p × p must include 2, 2, 2, 2, and 5.

Let's assign the prime factors to our p's. Since we have a 5 on our list of prime factors, we can give the 5 to one of our p's:

p: 5 p: p:

Since we have four 2's on our list, we can give each p a 2:

p: 5 × 2 p: 2 p: 2

But notice that we still have one 2 leftover. This 2 must be assigned to one of the p's:

p: 5 × 2 × 2 p: 2 p: 2

We must keep in mind that each p is equal in value to any other p. Therefore, all the p's must have exactly the same prime factorization (i.e. if one p has 5 as a prime factor, all p's must have 5 as a prime factor). We must add a 5 and a 2 to the 2nd and 3rd p's:

If p^3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

2 3 6 8 10

Prime factorize \(80 = 2^4 * 5\) If \(p^3\) has at least four 2s and a 5, it must have at least six 2s and three 5s (Every prime factor of \(p^3\) must have a power which is a multiple of 3). So p must have at least two 2s and a 5 as factors.

Minimum value of \(p = 2^2 * 5\) This gives us \((2+1)*(1+1) = 6\) distinct factors (at least)

Quote:

Can you explain the line in the bracket (Every prime factor of p^3 must have a power which is a multiple of 3)

Take any positive integer N.

Say \(N = 6 = 2*3\)

\(N^3 = 6^3 = (2^3 * 3^3)\)

Say \(N = 18 = 2 * 3^2\)

\(N^3 = 18^3 = (2^3 * 3^6)\)

Similarly, since p is a positive integer, it will be made up of some prime factors. When you cube it, every prime factor of p^3 will have a power of 3 or a multiple of 3.
_________________

If p^3 is divisible by 80, then the positive integer p must have at le [#permalink]

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23 Mar 2017, 17:43

Bunuel wrote:

Tough and Tricky questions: Factors.

If \(p^3\) is divisible by 80, then the positive integer p must have at least how many distinct factors?

(A) 2 (B) 3 (C) 6 (D) 8 (E) 10

OFFICIAL SOLUTION

The prime factorization of 80 is (2)(2)(2)(2)(5) = 2^4*5^1. Thus, \(p^3 = 2^4*5^1*x\), where x is some integer.

Assigning the factors of p 3 to the prime boxes of p will help us see what the factors of p could be.

The prime factors in ( ) above are factors not explicitly given for \(p^3\), but which must exist. We know that \(p^3\) is the cube of an integer, and must have “triples” of the prime factors of p. Since \(p^3\) has a factor of \(2^3\), p must have a factor of 2. The fact that \(p^3\) has an “extra” 2 and a 5 among its factors indicates that p has additional factors of 2 and 5.

If p is a multiple of (2)(2)(5) = 20, then at the very least p has 1, 2, 4, 5, 10, and 20 as factors. So we can conclude that p has at least 6 distinct factors.

Alternatively, we can use this shortcut for computing the number of factors: (2’s exponent + 1)(5’s exponent + 1) = (2 + 1)(1 + 1) = (3)(2) = 6.

The correct answer is C.

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If p^3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

(A) 2 (B) 3 (C) 6 (D) 8 (E) 10

Since p^3/80 = integer, we can say that the product of 80 and some integer n is equal to a perfect cube. In other words, 80n = p^3.

We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So let’s break down 80 into primes to help determine what extra prime factors we need to make 80n a perfect cube.

80 = 10 x 8 = 5 x 2 x 2 x 2 x 2 = 5^1 x 2^4

In order to make 80n a perfect cube, we need two more 2s, and two more 5s. Thus, the smallest perfect cube that is a multiple of 80 is 2^6 x 5^3.

To determine the least possible value of p, we can take the cube root of 2^6 x 5^3 and we have:

2^2 x 5^1

To determine the total number of factors, we add 1 to each exponent attached to each base and multiply those values together.

(2 + 1)(1 + 1) = 3 x 2 = 6 total factors.

Answer: C
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