We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

For some positive integer, the number n has 110n^3 positive integer divisors, including 1 and the number 100n^3. How many positive integer divisors does 81n^4 have?


Please Help Soon.

I would Greatly Appreciate It.

 Oct 9, 2019

1.Do the prime factorization of the number.Ex:a^p b^q c^r
2.Total no.of factors=(p+1)*(q+1)*(r+1)

Similarly here we have to do the prime factorization of the number 110n^3

prime factorization=110n^3=2*5*11*n^3
no.of factors=110 (already given)

Now we have to create a analogy in the system

no .of factors = 110=2*5*11

this means
1>respective powers of the prime factors will be one less than the above so they will be (1,4,10)
2>there will be total 3 factors ,so n much contain prime factors of one of the (2,5,11)

Now we have the full liberty to assign any power among above powers to any prime factors , but looking at the options below, we should not raise 5and 11 to the higher powers.

So considering 2^10 5^4 11^1 as one such combination because it will yield a number smaller in value and in horizon near to the answer choices.

So this means that 110n^3=2^10*5^4*11^1==> n=2^3*5

So 81*n^4=(3*n)^4==>2^12*5^4*3^4 (prime factorization)

Sum of factors ==>(12+1)*(4+1)*(4+1)==>325

 Oct 13, 2019

7 Online Users