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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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Re: If p ^3 is divisible by 80, then the positive integer p must [#permalink]
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01 Jun 2015, 20:06
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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27 Oct 2014, 18:45
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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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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28 Jun 2015, 20:57
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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28 Jun 2015, 22:18
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 formBut \(p^3 = 80x = 2^4*5*x\) 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 pi.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.



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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, 08:19
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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30 May 2016, 08:40
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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27 Dec 2016, 04:31
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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02 Feb 2017, 21:59
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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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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27 Mar 2017, 17:30
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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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 SOLUTIONp^3 divisible by 80 p^3 => factorising 80and writing in below format x=2 and y = 5 makes cube and then p= 2x2x5=20 222 2xx 5yy 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
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