divisors: find smallest number with 12 even divistors and...

[malick]

Hello.

Quite a while ago, I was asked to find the smallest positive integer with 12 positive even divisors and 6 positive odd divisors.

After trying for a while and not finding the answer, I read that the smallest number with 6 odd divisors is 3^2(5), and to get additionally 12 even, we must multiply by 2^2.
It is obviously true, but is there a trick to it or do you just do it the long way. Can someone explain?

 

[pka]

The first question for you is: “Do you understand how to count divisors?”
If \(\displaystyle N = 2^4 3^2 5^3\) then any divisor of N must look like \(\displaystyle 2^h 3^j 5^k ,\left\{ \begin{array}{l}
0 \le h \le 4 \\
0 \le j \le 2 \\
0 \le k \le 3 \\
\end{array} \right.\).
This means that h can have five values, j can have three values and k can have four values. So N has (5)(3)(4)=60 divisors. The number that are even means h cannot be zero so it has four values. Hence the number of even divisors is (4)(3)(4)=48.

 

[soroban]

Re: divisors: find smallest number with 12 even divistors an

Hello, malick!

Quite a while ago, I was asked to find the smallest positive integer
with 12 positive even divisors and 6 positive odd divisors.

There is a theorem for the number of divisors of \(\displaystyle N\).

If the prime factorization is: \(\displaystyle \,N \:=\:2^p\cdot3^q\cdot5^r\,\cdots\)

\(\displaystyle \;\;\)then the number of divisors is: \(\displaystyle \,d(N)\;=\;(p+1)(q+1)(r+1)\,\cdots\)

(Add 1 to each exponent and multiply) . . . This includes \(\displaystyle 1\) and \(\displaystyle N\) itself.


We have: \(\displaystyle \,d(N)\,=\,18\) . . . and \(\displaystyle \,18\:=\:2\cdot9\:=\:3\cdot6\:=\:2\cdot3\cdot3\)

\(\displaystyle \;\;\)so the exponents are: \(\displaystyle \,(1,8),\;(2,5),\;(1,2,2)\).


We have three candidates:
\(\displaystyle \;\;N\:=\:2^8\cdot3^1\:=\:768\)
\(\displaystyle \;\;N\:=\:2^5\cdot3^2\:=\:288\;\;\) *
\(\displaystyle \;\;N\:=\:2^2\cdot3^2\cdot5^1\:=\:180\)

The smallest of these is \(\displaystyle 180\).

which has 12 even divisors: 2, 4, 6, 10, 12, 18, 20, 30, 36, 60, 90, 180
. . . . . . .and 6 odd divisors: 1, 3, 5, 9, 15, 45.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
To get the smallest \(\displaystyle N\), place the larger exponents on the smaller prime factors.

 

[TchrWill]

Re: divisors: find smallest number with 12 even divistors an

Hello.

Quite a while ago, I was asked to find the smallest positive integer with 12 positive even divisors and 6 positive odd divisors.

After trying for a while and not finding the answer, I read that the smallest number with 6 odd divisors is 3^2(5), and to get additionally 12 even, we must multiply by 2^2.
It is obviously true, but is there a trick to it or do you just do it the long way. Can someone explain?

A number N of the form a^p(b^q)c^r... has a total number of factors of f(N) = (p + 1)(q + 1)(r + 1)...

If your number has 12 even divisors(factors) and 6 odd divisors (factors),for a total of 18 divisors, it must be of the form a^18, a(b^8), a^2(b^5), a^2(b^2)c, or a^2(b)c^2.

Assigning the primes from the smallest on up to the successive exponents, the smallest number of this form having 12 even divisors and 6 odd divisors would be 2^2(3^2)5 = 180 having divisors of 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90 and 180, 12 even and 6 odd.