IIT JEE , AIEEE Forum - Mathematics , Algebra - what are the total number of divisors and the sum of divisors of a given natural number?

rex_kj

what are the total number of divisors of 5400 and the sum of divisors which have 0 at their unit place?

sweet08

--------->for a given no. N=(P^m)(Q^n)(R^s)......whereP,Q,R are prime nos then no. of divisors=(m+1)(n+1)(s+1)

5400=3^3.2^3.5^2

so no of divisors=(3+1)(3+1)(2+1)=4*4*3=48

including the no. & 1

-------->sum of divisors=(1+P^1+P^2+P^3+........P^m)(1+Q^1+Q^2+..........Q^n)(1+R^1+R^2+..............R^s)

5400 sum of divisors==(1+3+3^2+3^3)(1+2+2^2+2^3)(1+5+5^2)

=40*15*31=18600