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If a factor of the number 2^33^55^27^4 is selected at random. What is

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If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 27 Nov 2019, 01:16
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A
B
C
D
E

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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 27 Nov 2019, 02:53
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Bunuel wrote:
If a factor of the number \(2^33^55^27^4\) is selected at random. What is the probability that it is odd?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{2}\)

D. \(\frac{6}{13}\)

E. \(\frac{11}{14}\)


Let N = \(2^33^55^27^4\)
Total factors of N = \((3 + 1)*(5 + 1)*(2 + 1)*(4 + 1) = 4*6*3*5\)

Odd factors of N is same as total factors of \(3^55^27^4\) = \((5 + 1)*(2 + 1)*(4 + 1) = 6*3*5\)

Probability of Odd factors = \(\frac{6*3*5}{4*6*3*5} = \frac{1}{4}\)

IMO Option B
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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 27 Nov 2019, 04:06
Total Factors: 4*6*3*5=360
Dismiss prime two from the set to get only odd factors: 6*3*5=90

So the prob to pick only odd is \(\frac{90}{360}=\frac{1}{4}\)

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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 27 Nov 2019, 07:59
From 2^3 * 3^5 * 5^2 * 7^4
Total factor = 3+5+2+4 (sum of every exponent) = 14
Odd factor = 5+2+4 (sum of odd exponent) = 11
So, prob. of odd factor = 11/14
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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 17 Dec 2019, 20:16
Bunuel chetan2u

Why aren't we accounting for repeats? There are 6 3's ...

So aren't we counting more than once..and hence need to account for it?i.e. divide the answer

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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 17 Dec 2019, 20:50
AdityaHongunti wrote:
Bunuel chetan2u

Why aren't we accounting for repeats? There are 6 3's ...

So aren't we counting more than once..and hence need to account for it?i.e. divide the answer

Posted from my mobile device



Hi

When you talk of 3^5 and we taking (5+1) or six 3s, it does not repeat anything
The 6 3s mean 3^0, 3^1, 3^2, 3^3, 3^4 and 3^5.

Solution
2^3*3^5*5^2*7^4
So odd factors will be given by 3^5*5^2*7^4........(5+1)(2+1)(4+1)=6*3*5
Total factors =(3+1)*6*3*5
Probability of odd factors =6*3*5/(4*6*3*5)=1/4

So what is happening here is the probability is just dependent on the power of even number.
Because each odd factor can be multiplied by 2^1 or 2^2 or 2^3, so 3 of them
Thus each odd factor leads to 3 more even factors, thus probability is 1*odd/(1*odd+3*even)=1/4
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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is  [#permalink]

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New post 17 Dec 2019, 20:53
chetan2u wrote:
AdityaHongunti wrote:
Bunuel chetan2u

Why aren't we accounting for repeats? There are 6 3's ...

So aren't we counting more than once..and hence need to account for it?i.e. divide the answer

Posted from my mobile device



Hi

When you talk of 3^5 and we taking (5+1) or six 3s, it does not repeat anything
The 6 3s mean 3^0, 3^1, 3^2, 3^3, 3^4 and 3^5.

Solution
2^3*3^5*5^2*7^4
So odd factors will be given by 3^5*5^2*7^4........(5+1)(2+1)(4+1)=6*3*5
Total factors =(3+1)*6*3*5
Probability of odd factors =6*3*5/(4*6*3*5)=1/4

So what is happening here is the probability is just dependent on the power of even number.
Because each odd factor can be multiplied by 2^1 or 2^2 or 2^3, so 3 of them
Thus each odd factor leads to 3 more even factors, thus probability is 1*odd/(1*odd+3*even)=1/4



Aah!! Silly mistake... I couldn't recognise that all the powers of 3 are in fact different numbers... Thank you so much !
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Re: If a factor of the number 2^33^55^27^4 is selected at random. What is   [#permalink] 17 Dec 2019, 20:53
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