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There are (3+1)*(4+1)*(5+1)=120 factors of 2^3*3^4*5^5.

The odd factors can be found where there is no 2 (e.g. 2^0*3^4*5^5). There are (0+1)*(4+1)*(5+1)=30 odd factors.

Consequently, there are (120-30)=90 even factors

Final answer is (C)

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Number : 2^3 * 3^4 * 5^5

Total Factors : 4*5*6 =120
Total Odd factors : 5*6 =30
Total Even factors : 3*5*6 =90 [Ignore 2^0]

Option C

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even no
4*5*6 ; 120
IMO E

How many factors of 2^3∗3^4∗5^5 are even numbers?

A. 20
B. 30
C. 90
D. 100
E. 120
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How many factors of \(2^3∗3^4∗5^5\) are even numbers?

A. 20
B. 30
C. 90
D. 100
E. 120

ALTERNATIVE approach first:
The factors of \(2^3∗3^4∗5^5\) can be written as
\(2^1, 2^2\) and \(2^3\) (only 2 OR its multiples) = 3
\(3^1, 3^2, 3^3\) and \(3^4\) (only 3 OR its multiples) = 4
\(5^1, 5^2, 5^3, 5^4\) and \(5^5\) (only 5 OR its multiples) = 5

Also, possible factors are the multiples of 3 and 5 = 4*5 = 20

So, total even factors are = Only 2's + Multiples of 2 and 3 + Multiples of 2 and 5 + Multiples of 2 and Multiples of 4 and 5
= 3 + 3*4 + 3*5 + 3*20
= 90

PS: The powers 2, 3 and 5 can take various values as follows:
Power of 2 = 3 ways (1 to 3; Can't take 0 as power 2 as \(2^0\) = 1 which is odd)
Power of 3 = 5 ways (0, 1 .. to .. 4)
Power of 5 = 6 ways (0, 1, .. to .. 6)
Hence total even factors = 3*5*6 = 90

-----------------------------------------------------------------------------------------------------

Number of factors for a number denoted as P^m * Q^n * R^o * ... = (m+1)*(n+1)*(o+1)* ..
Here we have only three prime factors, so
P = 2, Q = 3 and R = 5 and m = 3, n = 4 and o = 5

Hence total factors = (3+1)*(4+1)*(5+1) = 120
Out of these number of odd factors are = (5)*(6) = 30

Hence even factors are 120 - 30 = 90

Answer C.
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We are to determine the number of even factors of 2^3 * 3^4 * 5^5.
Total number of factors = (3+1)(4+1)(5+1)=4*5*6=120
Total number of odd factors=(4+1)(5+1)=5*6=30
Total number of even factors = 120-30 = 90.

The answer is C.
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The number of total factors of \(2^{3}∗3^{4}∗5^{5}\) is: (3+1)*(4+1)*(5+1) = 120
The number of odd factors of \(2^{3}∗3^{4}∗5^{5}\) is (4+1)*(5+1) = 30
=> The number of even factors of \(2^{3}∗3^{4}∗5^{5}\) is 120 - 30 = 90

Choice C
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Answer: C. 90
Explaination:
2^3*3^4*5^5 has (3+1)(4+1)(5+1)=120 factors
Odd factors of 2^3*3^4*5^5 do not contain any 2.
Number of odd factors= (4+1)(5+1)=30
Number of even factors= 120-30=90
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Ans: 90 (c)
ans=Total factors- odd factors
Total factor=4*5*6=120
Odd Factors= 1(2^0=1 possibility)*5*6=30
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Quote:
How many factors of 2^3∗3^4∗5^5 are even numbers?

A. 20
B. 30
C. 90
D. 100
E. 120

Total factors: 3+1*4+1*5+1=4*5*6=120
Odd factors (those w/o 2): 4+1*5+1=5*6=30
Even factors: 120-30=90

Ans (C)
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