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06 Feb 2017, 02:02
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If \(n=2^23^35^37\), how many factors of n are there?
A. 18
B. 36
C. 54
D. 96
E. 108
A. 18
B. 36
C. 54
D. 96
E. 108
Sirakri
06 Feb 2017, 02:27
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MathRevolution
If\(n=2^23^35^37\), how many factors of n are there?
A. 18
B. 36
C. 54
D. 96
E. 108
No of factors of \(a^mb^n\) is \((m+1)(n+1)\) where a,b are primes.A. 18
B. 36
C. 54
D. 96
E. 108
So, no of factors of \(n=2^23^35^37\):
\((2+1)(3+1)(3+1)(1+1)=3*4*4*2=96\)
D.
stonecold
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Joined: 12 Aug 2015
Last visit: 09 Apr 2024
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06 Feb 2017, 02:33
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Sirakri
MathRevolution
If\(n=2^23^35^37\), how many factors of n are there?
A. 18
B. 36
C. 54
D. 96
E. 108
No of factors of \(a^mb^n\) is \((m+1)(n+1)\) where a,b are primes.A. 18
B. 36
C. 54
D. 96
E. 108
So, no of factors of \(n=2^23^35^37\):
\((2+1)(3+1)(3+1)(1+1)=3*4*4*2=96\)
D.
->Distinct Primes.
Regards
Stone Cold
. 
