How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.  
 A: Let $p_i$, be primes in increasing order. Consider the prime factorization of $n$. 
$$ n = 2^{a_1} \times 3^{a_2} \times \prod p_i^{a_i}, $$
where $a_i$ are non-negative integers. For this question, let the indexing run over $i\geq 3$.
You are given that 
$$ (a_1 + 2) \times (a_2 +1 ) \times \prod (a_i + 1) = 28 $$
and
$$( a_1  + 1) \times (a_2 + 2) \times \prod(a_i + 1) = 30 $$
The factors of 28 are $1, 2, 4, 7, 14, 28$. The factors of 30 are $1, 2, 3, 5, 6, 10, 15, 30$.
A quick check shows that the possible values of $a_1$ are 2 or 5.    
If $a_1 = 2$, then we must have 
$$ \frac{ a_2 +1 } { a_2 + 2} = \frac{7}{10},$$
which has no positive integer solution.
If $a_1 = 5$, then we must have 
$$\frac{a_2 + 1} { a_2 + 2} = \frac{4}{5}, $$
which has a positive integer solution $a_2 = 3$. This implies that $\prod (a_i+1) = 1$. Hence, the number of factors of $6n$ is 
$$( a_1  + 2) \times (a_2 + 2) \times \prod(a_i + 1) = 7 \times 5 \times 1 = 35 $$
In this case, $n = 2^5 \times 3^3$.
A: If you multiply a number $n$ by any prime number, then its number of divisors is multiplied by $\frac{m+1}m$ where $p^{m-1}$ is the highest power of $p$ dividing$~n$. This is because the multiplicity of $p$ in divisors of $n$ can be any one of the $m$ values $0,1,\ldots,m-1$, but that multiplicity can be chosen among the $m+1$ values $0,1,\ldots,m$ for divisors of $pn$ (for prime numbers other than $p$, the same set of multiplicities is available in both cases).
So in the problem here let $k$ be the number of divisors of $n$ without factors $2$ or $3$, and $a-1,b-1$ the multiplicities of the primes $2$ and $3$ in$~n$, respectively. So $k,a,b$ are positive integers and $n$ has $kab$ divisors in all. We are given that $28=k(a+1)b$ and $30=ka(b+1)$, which implies that $\frac{28}{30}=\frac{14}{15}=\frac{(a+1)b}{a(b+1)}$. The factor $7$ in the numerator cannot come from $b$ (given that $b+1$ divides $~30$ and that $\frac{a+1}a\neq1$), so it must come from the factor $a+1$. Since $a$ divides $30$, the only possibility is $a=6$, which indeed gives a solution namely $(k,a,b)=(1,6,4)$. So we find $n=k2^{a-1}3^{b-1}=2^53^3$, and $6n=2^63^4$ has $k(a+1)(b+1)=35$ divisors.
A: If $2n$ has 28 divisors, then $2n$ is of the form $a^6.b$, where b has 4 divisors.  Note that if chose $2^13$ or $2^26$ here, then the third multiple would have more than $30$ divisors, since it would be $3*13* or $2*27$ divisors required.  
Since $3n$ does not have a divisor of the form $a^6$, we see that  $a=2$, and that n is at least $2^5$.  The reminder takes $b$ has $4$ divisors, and 3b has $5$, so $ b=3^3$. So the number is $864$, which has $6*4=24$ divisors, its second and third have $28$ and $30$ divisors.
Likewise $6n$ has 7*5 or $35$ divisors.  It is the $5184$.
One can also see that if $6n=2^a 3^b c$, then for $2n$ to have 28 divisors, and $3n$ to have 30 divisors, then $7|a-1$ and $5|b-1$.  This, and that $ab+b=28$ and $ab+a=30$, means that $a-b=2$, then $a=6, b=4$ is the only solution here.  
