Number of Multiples of $10^{44}$ that divide $ 10^{55}$ I have to find the number of multiples of $10^{44}$ that divide $ 10^{55}$ 
Multiples of $10^{44}$ would be  of the form $n.10^{44}$
Find how many values of n possible such that:  $n. 10^{44}$ divides $10^{55}$ 
or we can say, find those values of n such that: $n$ divides $10^{11}$ which indicates that the answer is $12\times12= 144$ . 
Please suggest if the answer and approach is correct.
 A: In reality, $10^{44}$ and $-10^{44}$ divide $10^{55}$, so it depends on if you want only the positive multiples, or all multiples.
If $n \cdot 10^{44}$ divides $10^{55}$, this is equivalent to $\frac{10^{55}}{n \cdot 10^{44}} = \frac{10^{11}}{n}$ being an integer, and this is equivalent to $n$ dividing $10^{11}$.
Now, you want to calculate the number of factors of $10^{11}$.  In general, for any positive number $a$, we can write it in its prime factorization as
$$a = p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}$$
where the $p_i$'s are all prime and distinct, and the $k_i$'s are all positive integers, then the number of positive factors is $(k_1 + 1)(k_2 + 1) \cdots (k_t + 1)$.  That is, any positive factor of $a$ must be of the form
$$p_1^{m_1} p_2^{m_2} \cdots p_t^{m_t}$$
where now I will allow the $m_i$'s to be 0 or positive, but $m_i \leq k_i$.  And, by the uniqueness of a prime factorization, two distinct choices of $m_i$'s lead to two distinct factors and vice versa.  Therefore, we simply need to count the number of ways to choose the $m_i$'s.  Clearly, there are $k_i + 1$ choices for $m_i$, which are $0, 1, 2, \ldots, k_i$.  Therefore we have the formula above.  In this case, we have $10^{11} = 2^{11} \cdot 5^{11}$ so there are $12 \cdot 12 = 144$ positive factors as you said.  But, if you allow negative multiples as well, then the answer is actually 2 times this, or 288.  That is, for each positive multiple you counted, there is also a corresponding negative multiple.
