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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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27 Nov 2019, 02:16
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35% (medium)

Question Stats:

73% (01:04) correct 27% (01:16) wrong based on 33 sessions

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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. $$\frac{11}{14}$$

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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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27 Nov 2019, 03:53
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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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27 Nov 2019, 05:06
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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Ans: 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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27 Nov 2019, 08:59
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
Ans E
Re: If a factor of the number 2^33^55^27^4 is selected at random. What is   [#permalink] 27 Nov 2019, 08:59
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