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2^3 * 3 *5^2

number of even factors = 3*(1+1)*(2+1)= 3*2*3=18
The first 3 comes from the 2 with power 1 and obove (2^1, 2^2, 2^3)

Answer: D
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Solution:



A factor of a number can either be even or odd.
Hence, we can write:
  • Total factors of a number = Even factors of the number + Odd factors of the number.
    • Total number of factors in \(2^3 * 3 * 5^2\) = \((3+1) * (1+1) * (2+1)\) =\(4*2*3\) = \(24\).
  • From our conceptual knowledge, any number written in its prime factorized form will have odd factors only if all the prime factors of the numbers are odd.
  • Similarly, here in this question, only \(3\) and \(5\) are such prime factors of the number.
    • Odd factors in \(2^3 * 3 * 5^2\) = number of factors in \(3 * 5^2\) = \((1+1) * (2+1) = 6\)
    Since the number of odd factors of \(2^3 * 3 * 5^2\) is \(6\), and we already have the total number of factors to be \(24\), the number of even factors of \(2^3 * 3 * 5^2 = 24-6 = 18\)
Hence, the correct answer is Option D.
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I tried to double-check when solving by using the manual way. 24 minus the factors
3, 5, 15, 25, 75
But this makes 19. I couldn't reconcile this for much too long.

Do not forget that 1 is a factor!
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philipssonicare
I tried to double-check when solving by using the manual way. 24 minus the factors
3, 5, 15, 25, 75
But this makes 19. I couldn't reconcile this for much too long.

Do not forget that 1 is a factor!

2^3* 3*5^2
factors of which can be written as 2^0 *(3*5^2) which is nothing but 1*(3*5^2)
2^1 *( 3*5^2)
2^2 *(3*5^2)
2^3*(3*5^2)
and for each of the above 3*5^2 will have (1+1)*(2+1) factors, so total number of factors is 6....
So for every (2^0, 2^1, 2^2 and 2^3) there are 6 combinations...
So for even factors at least one 2 should be present which gives 6*3 =18 factors

Now while calculating 6 factors for each 2 series we have already counted when each of them becomes 0
i.e Factors of 3*5^2 can be thought of (3^0*5^0=1, 3^0*5=5, 3^0*5^2=25, 3*5^0=3, 3*5^1=15, 3*5^2=75), so 6 factors are (1,5,25,3,15,75)
so 1 gets accounted in 6 factors of 3*5^2
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total factors = 4*2*3 = 24
Number of odd factors = 2*3=6
therefore even = 24-6=18
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EgmatQuantExpert

e-GMAT Question:



How many factors of the number \(2^3 * 3 * 5^2\) are even?

    A) 10
    B) 12
    C) 16
    D) 18
    E) 24


As we require even factors we need to ensure that the factors have at least one-two, put otherwise, we have to discount the possibility of \(2^0\)

Number of even factors = \(3 * 2 * 3 = 18\)

Option D
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