Question

A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer

Answer 1

As per my explanation to (b) above, the largest number of primes that could factor such a number is 4.

Note that  2,3,5 and 7   are the smallest primes, then use the reasoning from

(b) above. we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m, and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20 where k, l,m, and n ≥ 1 so this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors 2, 3, and 5 again we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have 2 *2 *5  = 20 and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining possibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20 clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648  or 2^4 * 3^3 = 432 and both are > 240.

Learn more about positive integers at

brainly.com/question/1367050

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