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

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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

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

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New post 01 Jun 2015, 20:06
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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)
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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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New post 27 Oct 2014, 18:45
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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


Let say p = 10, checking divisibility by 80

\(\frac{10 * 10 * 10}{80} = \frac{25}{2}\)

Numerator falling short of 2

So, lets say p = 20, again checking divisibility by 80

\(\frac{20*20*20}{80} = 100\)

20 is the least value of p for which \(p^3\) can be completely divided by 80

There are 6 distinct factors of 20 >> 1, 2, 4, 5, 10, 20

Answer = C

One more way:

\(20 = 2^2 * 5^1\)

Distinct factors = (2+1)*(1+1) = 3*2 = 6
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Factors and Divisibility [#permalink]

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New post 28 Jun 2015, 20:25
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
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If p^3 is divisible by 80, then the positive integer p must have at le [#permalink]

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New post 28 Jun 2015, 20:57
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abhisheknandy08 wrote:
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
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Re: Factors and Divisibility [#permalink]

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New post 28 Jun 2015, 22:18
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abhisheknandy08 wrote:
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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New post 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.
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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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New post 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)

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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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New post 30 May 2016, 08:40
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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


\(p^3\) = 80m, where m is any integer.

Now here key point is that 80m is a perfect cube. [ We know this because it is given that p is a positive integer]

Now the question says "at least".

let us see how 80m can be a perfect cube.

80m = 8 * 2 *5 *m = \(2^3\) * 2 * 5 * m

So we need to multiply "2" by \(2^2\), so that we get \(2^3\)
We also need to multiply "5" by \(5^2\), so that we get \(5^3\)

so m is \(2^2\) * \(5^2\)

With above value of m, p becomes (at least) 2*2*5 = \(2^2\) * 5

Distinct factors (Power of First term +1) (Power of Second term +1) [ You need to know this formula]

(2+1)(1+1) = 6

C is the answer.
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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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New post 27 Dec 2016, 04:31
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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:

p: 5 × 2 × 2 = 20
p: 5 × 2 × 2 = 20
p: 5 × 2 × 2 = 20

We conclude that p must be at least 20 for p^3 to be divisible by 80. So, let's count how many factors 20, or p, has:

1 × 20
2 × 10
4 × 5

20 has 6 factors. If p must be at least 20, p has at least 6 distinct factors.

The correct answer is 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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New post 02 Feb 2017, 21:59
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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)


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

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

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New post 27 Mar 2017, 17:30
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Quote:

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

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New post 28 Jul 2017, 12:29
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



(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.


SO why isn't it 2^12?
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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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New post 12 Sep 2017, 00:08
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


SOLUTION
p^3 divisible by 80

p^3 => factorising 80and writing in below format x=2 and y = 5 makes cube and then p= 2x2x5=20
2-2-2
2-x-x
5-y-y

20= 2^2 . 5^1
total factors = (2+1) ( 1+1) = 3x2= 6

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

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New post 10 Jan 2018, 13:29
Given \(P^3\) divisible by 80
=> \(P^3\) = \(2^4 * 5 * m\) (where m is any integer)
=> to make P as integer, min value of m = \(2^2 * 5^2\)
=> min value of \(P^3\) = \(2^4 * 5 * 2^2 * 5^2\)
=> min value of P = \(2^2 * 5\)
mininum number of factors of P = (2+1)(1+1) = 6 => (C)
If p^3 is divisible by 80, then the positive integer p must have at le   [#permalink] 10 Jan 2018, 13:29
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