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22 Jan 2021, 07:45
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (01:40) correct
50%
(01:50)
wrong
based on 112
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive factors does \(N = 2^3*3^2*5^4\) have that end with zero?
A. 24
B. 36
C. 48
D. 50
E. 60
A. 24
B. 36
C. 48
D. 50
E. 60
22 Jan 2021, 07:59
Kudos
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Bunuel
\(N = 2^3*3^2*5^4\)
FOr having zero at the end of a number, the result must be a multiple of 10 therefore
\(N = (10)*2^2*3^2*5^3\)
CONCEPT:
Number of FactorsFor a Number \(N=a^p*b^q*c^r*... \)where a, b, c etc are distinct prime factors of n and p, q r are their respective exponents
The number of factors of \(N = (p+1)*(q+1)*(r+1)*...\) and so on
So keeping 10 aside, the total factors of \(\frac{N}{10} = 2^2*3^2*5^3\) will be (2+1)*(2+1)*(3+1) = 36
i.e. N/10 has 36 factors
so N will have 36 factors that are multiples of 10
Answer: Option B
General Discussion
22 Jan 2021, 18:30
Kudos
Bookmarks
Bunuel
To end with a 0, we need at least one factor of two and at least one factor of five (to create a factor of 10). Given N, we are allowed to pick up to 3 multiples of two, up to 2 multiples of three, and up to 4 multiples of five.
We need at least one multiple of two, so we can choose from 1, 2, or 3 multiples of twos (3 options).
Similarly, we have 4 choices for fives.
For three we can choose 0, 1, or 2 multiples so that is 3 choices.
Putting everything together we have 3 * 4 * 3 = 36 choices.
Ans: B
