Find least possible number of factors Number $A$ has $24$ factors. Number $A\cdot B$ has $105$ factors. Find least possible number of factors of $B$. I have tried. But there seems to be no general approach.. The answer given is $12$.
 A: Let $n=AB=\prod_{j=1}^{k}p_j^{n_j},$ then the number of divisors is
$\sigma(n)=\prod_{j=1}^k(n_j+1)=105.$ So $k=3.$
Let $A=\prod_{j=1}^{3}p_j^{a_j},\ B=\prod_{j=1}^{3}p_j^{b_j}.$ 
Then $\sigma(A)=(a_1+1)(a_2+1)(a_3+1)=24$ and $\sigma(B)=(b_1+1)(b_2+1)(b_3+1).$
We have $a_j+b_j=n_j$ (and so $0\leq a_j,b_j\leq n_j$) and
without loss of generality
$(n_1+1,n_2+1,n_3+1)=(3,5,7).$ So
$$(a_1,a_2,a_3)\in \{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(0,3,5),(1,1,5)\}.$$
This gives the following values for 
$$\sigma(B)=[(n_1+1)-a_1][(n_2+1)-a_2][(n_3+1)-a_3]=$$
$$(3-a_1)(5-a_2)(7-a_3)\in \{24,20,16,{\bf 12},{\bf 12},16\}.$$
$\small{\text{(hope not to forget something)}}$
A: A partial answer :
The number $2^2\times3^4\times5^6$ has $105$ factors, the number $3^3\times5^5$ has $24$ factors and the number $2^2\times3\times5$ has $12$ factors. So, there is an example,
 such that $B$ has $12$ factors. 
A: Let $A = p_1 ^ {k_1} \times p_2 ^ {k_2} ... p_n ^ {k_n}$
So $24 = (k_1 + 1) \times (k_2 + 1) ... (k_n +1)$
Let $B = p_1 ^ {l_1} \times p_2 ^ {l_2} ... p_n ^ {l_n} \times q_1 ^ {m_1} \times q_2 ^ {m_2} ... q_v ^ {m_v} ...$ allowing for some $l_i$'s to be $0$ if necessary. 
Then $105 = ({k_1} + {l_1} + 1) \times ({k_2} + {l_2} + 1) ... ({k_n} + {l_n} + 1) \times ({m_1} + 1) .. ({m_v} + 1)$
But 105 = 3 x 5 x 7 ... So n <= 3
Case 1:    n=1,     $  A = p^{23}$ 
... most economical $AB = p ^ {34} \times q ^ 2  $ with B having 12 x 3 = 36 factors
Case 2: n=2
i.    $A = {p_1} ^ {11} \times {p_2} ^ 1$   
... most economical $AB = {p_1} ^ {14} \times {p_2} ^ 6$ with B having 4 x 6 = 24 factors
ii.    $A = {p_1} ^ 7 \times {p_2} ^ 2$   
... most economical $AB = {p_1} ^ {14} x {p_2} ^ 6$ with B having 8 x 5 = 40 factors
iii.    $A = {p_1} ^ 5 \times {p_2} ^ 3$   
... most economical $AB = {p_1} ^ 6 \times {p_2} ^ 4 \times q ^ 2$ with B having 12 factors
Case 3 n=3
i.    $ A = {p_1} ^ 5 \times {p_2} ^ 1 \times {p_3}$  
... most economical $AB = {p_1} ^ 6 \times {p_2} ^ 2 \times {p_3} ^ 4$ with B having 16 factors
ii.     $A = {p_1} ^ 3 \times {p_2} ^ 2 \times {p_3}$  
... most economical $AB = {p_1} ^ 4 \times {p_2} ^ 2 \times {p_3} ^ 6$ with B having 12 factors also
This search is exhaustive which completes the proof.
A: The general approach is this (example given with $24$ and $105$):


*

*$24=6\cdot4\implies{p^{6-1}}\cdot{q^{4-1}}$ has $24$ divisors

*$105=7\cdot5\cdot3\implies{p^{7-1}}\cdot{q^{5-1}}\cdot{r^{3-1}}$ has $105$ divisors

*$\dfrac{{p^{7-1}}\cdot{q^{5-1}}\cdot{r^{3-1}}}{{p^{6-1}}\cdot{q^{4-1}}}={p^{2-1}}\cdot{q^{2-1}}\cdot{r^{3-1}}$ has $2\cdot2\cdot3=12$ divisors
Any distinct primes $[p,q,r]$ that you choose will gave you the same number of divisors.

Each one of the following is a valid combination of A and AB:
         |  105  |  3*35 |  5*21 |  7*15 | 3*5*7
---------|-------|-------|-------|-------|-------
 24      |       |       |   X   |   X   |   X  
---------|-------|-------|-------|-------|-------
 2*12    |   X   |       |       |       |   X  
---------|-------|-------|-------|-------|-------
 3*8     |   X   |       |       |       |   X  
---------|-------|-------|-------|-------|-------
 4*6     |   X   |   X   |       |       |   V  
---------|-------|-------|-------|-------|-------
 2*2*6   |   X   |   X   |   X   |   X   |      
---------|-------|-------|-------|-------|-------
 2*3*4   |   X   |   X   |   X   |   X   |      
---------|-------|-------|-------|-------|-------
 2*2*2*3 |   X   |   X   |   X   |   X   |   X  

In the combinations marked X, the value of B will not be integer.
In general, you should try each of the other combinations in order to find out which one yields a value of B with the lowest number of divisors (the combination marked V is the one that I tested).
