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Number of Multiples of $10^{44}$ that divide $ 10^{55}$

I want to find out the number of multiples of $10^{44}$ that divide $10^{55}$ from the following options.

$(a)\ 11\ (b)\ 12\ (c)\ 121\ (d)\ 144$. I don't know which option is correct and please explain the method.

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marked as duplicate by Marvis, Jason DeVito, Dylan Moreland, The Chaz 2.0, mixedmath May 20 '12 at 5:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is a duplicate of a previous question. – Antonio Vargas May 20 '12 at 3:26
up vote 4 down vote accepted

Any multiple of $10^{44}$ is of the form $10^{44} a$, where $a \in \mathbb{Z}$. We want $10^{44} a \vert 10^{55}$. Equivalently, we want $a \vert 10^{11}$. So all you need is to count the number of divisors of $10^{11}$. Can you complete it from here?

Move your mouse over the gray area below for another hint.

In general, if you know that prime factorization of $m$, i.e. say $m= p_1^{\alpha}p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then by a simple counting argument the number of divisors of $m$ is given by (why?) $$d(m) = (1+\alpha_1)(1+\alpha_2) \cdots (1+\alpha_k)$$

Move the mouse over the gray area below for the complete answer.

Hence, all you need to get the prime decomposition of $10^{11}$. The prime decomposition of $10^{11}$ is $2^{11} \times 5^{11}$. Hence, the number of divisors of $10^{11}$ is given by $(1+11) \times (1+11) = 144$. Hence, the number of multiples of $10^{44}$ that divide $10^{55}$ is $144$.

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ok! number of divisors of $10^{11}=2^{11}.5^{11}$ is $144$. I am using $\tau{(p^{\alpha})}=(\alpha +1)$. – Kns May 20 '12 at 3:26
@KunjanShah Yes. you are correct. – user17762 May 20 '12 at 3:27
Thanks a lot Marvis. – Kns May 20 '12 at 3:28

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