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Bunuel
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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
Ans E
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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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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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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 =\(\frac{6*3*5}{(4*6*3*5)}=\frac{1}{4}\)

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 \(\frac{1*odd}{(1*odd+3*even)}=\frac{1}{4}\)
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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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Why did you guys add 1 to each number if the individual number of the factors?
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