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p is the smallest perfect cube greater than 1

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p is the smallest perfect cube greater than 1  [#permalink]

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New post Updated on: 07 Aug 2018, 03:20
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Question 2 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors

\(p\) is the smallest perfect cube greater than 1 such that the difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2. If \(Z = 1*2*3*…*p\), then the total number of factors of \(Z\) is

A. 5
B. 8
C. 14
D. 96
E. 192

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Originally posted by EgmatQuantExpert on 04 May 2015, 04:59.
Last edited by EgmatQuantExpert on 07 Aug 2018, 03:20, edited 4 times in total.
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post Updated on: 04 May 2015, 06:31
3
The next perfect cube greater than 1 is 8 , so p' is 8.

8×8=64

6-4=2


hence,

Z=8!
1×2×3×4×5×6×7×8=1×2×2×2×2×2×2×2×3×3×5×7

Number of factors=2×8×3×2=96
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Originally posted by Lucky2783 on 04 May 2015, 05:13.
Last edited by Lucky2783 on 04 May 2015, 06:31, edited 1 time in total.
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 04 May 2015, 06:14
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Lucky2783 wrote:
The next perfect cube greater than 1 is 8 , so p' is 8.

8×8=64

6-4=2


hence,

Z=8!
1×2×3×4×5×6×7×8=1×2×2×2×2×2×2×2×3×3×5×7

Number of factors=2×8×3×2×2=192



I think you multiplied one 2 more. The answer is - 8x3x2x2 = 96.
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 04 May 2015, 06:39
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first perfect cube greater than 1 is 8 and as it so happens it is the one that matches our conditions, thus our Z is 8!
\(8! = 2*3*4*5*6*7*8 = 2^7*3^2*5^1*7^1\)
the amount of multiples is easily found with the formula: \((7+1)*(2+1)*(1+1)*(1+1) = 8*3*2*2=96\)
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 04 May 2015, 18:32
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The first perfect cube after greater than 1 is 8.

\(8^2 = 64\)

The difference between the digits = 6-4 = 2
Hence, p = 8.

\(Z= 8! = 2^7 * 3^2 * 5 * 7\)
Total no of factors of Z = (7+1)*(3+1)*(1+1)*(1+1) = 96
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 05 May 2015, 22:01
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Thank you all for your attempts. Here is the solution for this question.

Given: \(p\) is the smallest perfect cube greater than 1 such that difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2.

We are also given that \(Z = 1*2*3*…*p\)

The question asks us to find the total number of factors of \(Z\).

Approach: To find the total number of factors of \(Z\), we need to know the power of every prime factor of \(Z\). Let’s say Z is of the form \(Z = a^m * b^n * …\). where \(a, b\),… are primes, then the total number of factors of \(Z\) will be \((m+1)*(n+1)…\)

Therefore to be able to express \(Z\) in its prime factorized form, we need the value of \(p\).

We will try to find the value of \(p\) based on the constraints imposed on it:

    a. \(p\) is a perfect cube greater than 1
    b. difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2

Working Out: \(p\) is a perfect cube greater than 1.

Possible values of \(p\) are 8, 27, 64…

When \(p = 8\), \(p^2 = 64\).

Notice that this satisfies the second condition too (difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2). So we need not check other values for \(p\).

Therefore \(p = 8\)

This essentially means \(Z = 1*2*3*4*5*6*7*8 = 2 * 3 * 2^2 * 5 * 2*3 * 7 * 2^3 = 2^7 * 3^2 * 5 * 7\)

Therefore total number of factors of \(Z\) = \((7+1)*(2+1)*(1+1)*(1+1) = 96\)

Correct Answer: Option D


Here is another question that tests your conceptual understanding of primes and factors.
x-y-are-integers-find-the-number-of-even-factors-of-4x-197375.html


Hope this helps. :)

Regards,
Krishna
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 05 May 2015, 22:12
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Q : p is the smallest perfect cube greater than 1 such that the difference between the tens digit of p2 and the units digit of p2 is 2. If Z=1∗2∗3∗…∗p, then the total number of factors of Z is

smallest perfect cube > 1 ; p ~ (8,27,64 ....)
\(p^{2} --> (64,729,64*64 ....)\)
64 = difference between the tens digit and the units digit = 2 ; implies p=8

p=8

Z=1∗2∗3∗…∗8
\(Z= 2^{7}∗3^{2}∗5^{1}∗7^{1}\)
factors = (7+1)(2+1)(1+1)(1+1) = 96

Ans D
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 05 May 2015, 22:23
Lucky2783 wrote:
The next perfect cube greater than 1 is 8 , so p' is 8.


Although you arrived at the right answer, I think your interpretation of the question was incorrect here. Please refer to the solution and let me know if that clears things up.

Hope this helps. :)

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Krishna
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 18 Aug 2015, 13:48
What's the meaning behind this: (7+1)(2+1)(1+1)(1+1) = 96???
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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New post 18 Aug 2015, 13:54
immanl wrote:
What's the meaning behind this: (7+1)(2+1)(1+1)(1+1) = 96???


Finding the Number of Factors of an Integer:

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 \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Check for more here: math-number-theory-88376.html
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