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Let n be the smallest positive integer such that n is divisible by 20,

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Bunuel
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Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   31 Mar 2019, 20:58
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Let n be the smallest positive integer such that n is divisible by 20, n^2 is a perfect cube, and n^3 is a perfect square. What is the number of digits of n?

A. 3
B. 4
C. 5
D. 6
E. 7
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E
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jfranciscocuencag
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Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   02 Apr 2019, 15:14
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It took me some time to figure it out.

Factorize \(20\) --- \((5^1)(2^2)\)

We have to equal the exponents and look for the smallest number that fills the rules written above.

So first we try with \((10)^2\) --- \((5^2)(2^2)\), then with \((10)^3\) and so on.

\((10^6)^2 = (10)^12 = (10^4)(10^4)(10^4)\)

\((10^6)^3 = (10)^18 = (10^9)(10^9)\)

So \((10^6)\) It has 7 digits.

E
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Re: Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   04 Apr 2019, 18:20
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Bunuel wrote:
Let n be the smallest positive integer such that n is divisible by 20, n^2 is a perfect cube, and n^3 is a perfect square. What is the number of digits of n?

A. 3
B. 4
C. 5
D. 6
E. 7


If n is a positive integer and n^2 is a perfect cube, then n itself has to be a perfect cube. That is, we can write n^2 = m^3 for some integer m. So n = √(m^3) = (√m)^3. We see that √m must be an integer since n is an integer. Therefore, we see that n is a perfect cube. Using the same argument, we can conclude that n is a also a perfect square since n^3 is a perfect square.

Since n is a perfect square and a perfect cube, n is a perfect 6th power. Since n is also divisible by 20 = 2^2 x 5, the smallest positive value of n is 2^6 x 5^6 = 10^6, which is 1 followed by 6 zeros. Therefore, n has 7 digits.

Answer: E
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Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   13 Nov 2019, 07:45
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Bunuel wrote:
Let n be the smallest positive integer such that n is divisible by 20, n^2 is a perfect cube, and n^3 is a perfect square. What is the number of digits of n?

A. 3
B. 4
C. 5
D. 6
E. 7


still confuse :cry: with others explanation. Better if any other explanation. :) Bunuel help me to figure it out.
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Re: Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   06 Jan 2021, 00:58
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Bunuel wrote:
Let n be the smallest positive integer such that n is divisible by 20, n^2 is a perfect cube, and n^3 is a perfect square. What is the number of digits of n?

A. 3
B. 4
C. 5
D. 6
E. 7



For a certain number to be both a cube and perfect square, that is power of 2 and 3, the power has to be a multiple of 2*3 or 6.
So \(n=a^6\), where a is an integer.
If we did not have any restriction, the answer would be 1.

But here n is a multiple of 20 => 20x
So, \(20x=a^6......2^2*5*x=a^6\)
For a to be the smallest positive integer possible, x should be \(2^4*5^5\)
\(n=2^2*5*2^4*5^5=2^6*5^6=(2*5)^6=10^6=1,000,000\)

Thus 7 digits.

E
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Re: Let n be the smallest positive integer such that n is divisible by 20, [#permalink] New post   14 Jun 2023, 18:53
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Re: Let n be the smallest positive integer such that n is divisible by 20, [#permalink]
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