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If m=a^4b^2c^5d is the prime factorization of m, what is the number of
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01 Jul 2021, 04:29
01 Jul 2021, 04:29
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If \(m=a^4b^2c^5d\) is the prime factorization of m, what is the number of positive divisors of m ?
A. 12
B. 40
C. 80
D. 90
E. 180
A. 12
B. 40
C. 80
D. 90
E. 180
Re: If m=a^4b^2c^5d is the prime factorization of m, what is the number of
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01 Jul 2021, 05:12
01 Jul 2021, 05:12
Solution:
m=\(a^4b^2c^5d\)
In order to find the total number of positive zeros, add 1 to the power of the numbers and multiply
E.g. 6=2*3
Thus, 2 & 3 have a power of 1
Add 1 to both of them and multiply
(1+1)(1+1)=4
You can try finding the individual factors(1,6,2 & 3) the total number of factors will be 4.
Similarly, for the above adding 1 to all the power gives us:
(4+1)(2+1)(5+1)(1+1)=4*3*5*2=180
IMO E
Hope this helps!
m=\(a^4b^2c^5d\)
In order to find the total number of positive zeros, add 1 to the power of the numbers and multiply
E.g. 6=2*3
Thus, 2 & 3 have a power of 1
Add 1 to both of them and multiply
(1+1)(1+1)=4
You can try finding the individual factors(1,6,2 & 3) the total number of factors will be 4.
Similarly, for the above adding 1 to all the power gives us:
(4+1)(2+1)(5+1)(1+1)=4*3*5*2=180
IMO E
Hope this helps!
gmatclubot
Re: If m=a^4b^2c^5d is the prime factorization of m, what is the number of [#permalink]
01 Jul 2021, 05:12
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