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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.

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The solution seems pretty right to me. –  Tomarinator May 18 '12 at 14:38
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@preeti: In the second line, don't you mean 'Multiples of $10^{44}$ would be of the form $n\cdot 10^{44}$'? (Rather than the number of multiples as you've said.) –  Tara B May 18 '12 at 14:47
    
@TaraB: Thanks, you are right. –  preeti May 18 '12 at 14:49
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preeti, the original tags group-theory and linear-algebra were removed because this question isn't directly related to either of those topics. The current elementary-number-theory tag is much more appropriate. –  Antonio Vargas May 18 '12 at 15:35

1 Answer 1

up vote 3 down vote accepted

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.

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Many thanks for the informative reply. –  preeti May 18 '12 at 16:39

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