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Bunuel
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If N = 2^2 * 3^3 * 5^5, what is the total number of odd and even 29 Nov 2022, 12:30
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C
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25% (medium)

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73% (01:16) correct 28% (02:20) wrong based on 40 sessions

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If \(N = 2^2 * 3^3 * 5^5\), what is the total number of odd and even factors of N, respectively ?

(A) 24, 96
(B) 15, 30
(C) 24, 48
(D) 15, 96
(E) 15, 48
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C
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If N = 2^2 * 3^3 * 5^5, what is the total number of odd and even 29 Nov 2022, 12:48
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Number of odd Factors
= (3+1)*(5+1)
=24
Number of even factors
= 2*(3+1)*(5+1)
=48
Option C is the correct answer

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If N = 2^2 * 3^3 * 5^5, what is the total number of odd and even 29 Nov 2022, 23:18
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Alamin0160
Number of odd Factors
= (3+1)*(5+1)
=24
Number of even factors
= 2*(3+1)*(5+1)
=48
Option C is the correct answer

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Can you please explain this with a more detail.
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divyadna
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If N = 2^2 * 3^3 * 5^5, what is the total number of odd and even 29 Nov 2022, 23:41
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Total number of factors = (2+1)*(3+1)*(5+1)= 3*4*6=72

The only option which gives a total of 72 is option C i.e. 24+48=72.
Option C should be the answer

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If N = 2^2 * 3^3 * 5^5, what is the total number of odd and even 30 Nov 2022, 01:47
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Bunuel
If \(N = 2^2 * 3^3 * 5^5\), what is the total number of odd and even factors of N, respectively ?

(A) 24, 96
(B) 15, 30
(C) 24, 48
(D) 15, 96
(E) 15, 48
Show more


For odd no.of factors one should remove 2^2, and then calculate the no.of factors i.e N=3^3 * 5^5 will have (3+1)(5+1)=4*6=24 (no.of odd factors)
For even no.of factors one should include 2 only once ir-respective of the power of 2, i.e N=2^1 * 3^3 * 5^5 will have (1+1)(3+1)(5+1)=2*4*6=48 (no.of odd factors)
Hence, C is the correct choice.
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