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?

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Guest Oct 9, 2019

#1**+2 **

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

SVS2652 Oct 13, 2019