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# (a) Let \$m\$ be a positive integer such that \$m\$ has exactly 5 positive divisors. How many distinct prime factors does \$m\$ have?

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(a) Let \$m\$ be a positive integer such that \$m\$ has exactly 5 positive divisors. How many distinct prime factors does \$m\$ have?

(b)Let \$m\$ and \$n\$ be positive integers such that \$m\$ has exactly 5 positive divisors, \$n\$ has exactly 6 positive divisors, and \$mn\$ has exactly 14 positive divisors. How many distinct prime factors does \$mn\$ have?

Mellie  Jul 1, 2015

#2
+85809
+10

Here's my take on (b)

As per part (a), if m has 5 distinct positive divisors, it only has one prime factor - and that to the 4th power.  Let's call this prime "p."

And m*n has 14 positive divisors....... and,  if n has six distinct positve divisors, it must have "p" as one of its prime factors as well.....and it must be to the second power......this would also mean that "n" would have a second prime as a factor....call it "q"....and it would have to be to the first power......

Proof......m*n   =   [p^4]* [p^2 * q]   =  p^6*q.......note that m has (4 + 1) = 5 positive divisors....and  n has (2 + 1)(1 + 1) = (3)(2)  = 6 positive divisors and m*n has (6 + 1)(1 + 1) = (7)(2) = 14 positive divisors...therefore....m*n is composed of two distinct primes.....

CPhill  Jul 1, 2015
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#1
+85809
+10

Here's (a)

If  m has exactly 5 positive divisors.....there can only be one prime factor....to see why....

Let's suppose m can be prime factored as.....a^p * b^q , ......the number of possible divisors of m is  (p +1)(q + 1)  and we assume that p and q ≥ 1.......then the number of possible divisors cannot be 5 because there are no two integers  p, q ≥ 1, such that (p +1)(q + 1) = 5 .....

Similarly, if m factored as some number of primes greater than two, the least number of possible divisors would be eight... (Example......if a, b and c were primes each to the first power, the least number of divisors would be ....  2 * 2 * 2  = 8 )

Therefore, if m has 5 positive divisors, then it can only have one prime factor - and this has to be to the 4th power......( Example  ...  2^4  = 16 ....  and it has 5 positive divisors......1, 2, 4, 8, 16......)

CPhill  Jul 1, 2015
#2
+85809
+10

Here's my take on (b)

As per part (a), if m has 5 distinct positive divisors, it only has one prime factor - and that to the 4th power.  Let's call this prime "p."

And m*n has 14 positive divisors....... and,  if n has six distinct positve divisors, it must have "p" as one of its prime factors as well.....and it must be to the second power......this would also mean that "n" would have a second prime as a factor....call it "q"....and it would have to be to the first power......

Proof......m*n   =   [p^4]* [p^2 * q]   =  p^6*q.......note that m has (4 + 1) = 5 positive divisors....and  n has (2 + 1)(1 + 1) = (3)(2)  = 6 positive divisors and m*n has (6 + 1)(1 + 1) = (7)(2) = 14 positive divisors...therefore....m*n is composed of two distinct primes.....

CPhill  Jul 1, 2015
#3
+92221
+5

You are on a roll today Chris.  Your logic is irrefutable

(That's what I think anyway)      LOL

Melody  Jul 2, 2015

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