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

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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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27 Oct 2014, 07:51
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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
[Reveal] Spoiler: OA

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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 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

One more way:

$$20 = 2^2 * 5^1$$

Distinct factors = (2+1)*(1+1) = 3*2 = 6
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Re: If p ^3 is divisible by 80, then the positive integer p must [#permalink]

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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 13 Mar 2013 Posts: 175 Location: United States Concentration: Leadership, Technology GPA: 3.5 WE: Engineering (Telecommunications) Factors and Divisibility [#permalink] ### Show Tags 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 _________________ Regards , Math Expert Joined: 02 Aug 2009 Posts: 5732 If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 28 Jun 2015, 20:57 3 This post received KUDOS Expert's post 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 _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor SVP Joined: 08 Jul 2010 Posts: 2017 Location: India GMAT: INSIGHT WE: Education (Education) Re: Factors and Divisibility [#permalink] ### Show Tags 28 Jun 2015, 22:18 4 This post received KUDOS Expert's post 4 This post was BOOKMARKED 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 _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html 22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Intern Joined: 06 Mar 2015 Posts: 29 Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 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. Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3366 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 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.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Senior Manager Joined: 18 Jan 2010 Posts: 256 Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 30 May 2016, 08:40 1 This post received KUDOS 1 This post was BOOKMARKED 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. Director Joined: 26 Oct 2016 Posts: 682 Location: United States Concentration: Marketing, International Business Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 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: 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. _________________ Thanks & Regards, Anaira Mitch Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8002 Location: Pune, India Re: If p^3 is divisible by 80, then the positive integer p must have at le [#permalink] ### Show Tags 02 Feb 2017, 21:59 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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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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.

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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 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.

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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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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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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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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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