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Understand the problem

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

Source of the problem
American Mathematics Competition
Number Theory

8/10

Suggested Book

Elementary Number Theory by David M. Burton

Check the problem out…give its statement a thorough read. Might appear a bit daunting on the first couple of reads. Think for some time, you could be on to something without any help whatsoever !

Okay, now let’s think about what our first thoughts could be, on the problem. It’s definitely about the n in the problem, which acts as our unknown here.  Can you somehow try finding the n ? Let’s take the first step in that direction. How could we prime factorize 110 ? That’s easy 110 = 2.5.11. Could you take things from hereon to find more about the n ?

However interestingly the problem says, the number 110. (n^3)  has 110 factors. Just as we saw, 110. (n^3) = 2.5.11.(n^3) Now, let’s use some basic number theoretic knowledge here. How many divisors would 110. (n^3) have then ?  If n=1 Clearly it would have, (1+1). (1+1). (1+1) = 8 divisors.  So see, that’s the idea isn’t it ? Pretty much of plug and play. We actually get to control how many divisors the number has, once we adjust (n^3).  Now you could try some advances…

Okay, so as we just understood we need to achieve a count of 110 divisors.  If we have 110.(n^3) = 2^(10). 5^(4). 11 which actually conforms to :  (10+1).(4+1).(1+1) = 11.5.2 = 110  So, that implies :   (n^3) = 2^(9). 5^(3), which means, n = 2^(3). 5 Now that we have found out n…the rest dosen’t seem really a big deal. You could do it…try !

Well, it’s pretty straightforward now.  Let’s call 81.(n^4) equal to some X. First let’s prime factorize 81. That would be 81 = (3^4). So, finally X = (3^4). (2^12). (5^4) How many divisors does that make ? Yes, (4+1).(12+1). (4+1) = 13.25 = 325.

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