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# p is the smallest perfect cube greater than 1

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Joined: 04 Jan 2015
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p is the smallest perfect cube greater than 1  [#permalink]

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Updated on: 07 Aug 2018, 04:20
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Difficulty:

65% (hard)

Question Stats:

64% (01:54) correct 36% (01:46) wrong based on 299 sessions

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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, 05:59.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:20, edited 4 times in total.
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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Updated on: 04 May 2015, 07:31
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, 06:13.
Last edited by Lucky2783 on 04 May 2015, 07:31, edited 1 time in total.
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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04 May 2015, 07:14
1
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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04 May 2015, 07:39
3
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: $$(7+1)*(2+1)*(1+1)*(1+1) = 8*3*2*2=96$$
D
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Re: p is the smallest perfect cube greater than 1  [#permalink]

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04 May 2015, 19:32
1
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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05 May 2015, 23:01
1
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 $$Z$$ = $$(7+1)*(2+1)*(1+1)*(1+1) = 96$$

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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05 May 2015, 23:12
1
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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05 May 2015, 23:23
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.

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

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18 Aug 2015, 14:48
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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18 Aug 2015, 14:54
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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Re: p is the smallest perfect cube greater than 1  [#permalink]

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