MHB What Is the Smallest Integer $a$ for $\frac{1}{640}=\frac{a}{10^b}$?

[anemone]

Here is this week's POTW:

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Find the smallest value of $a$ such that $\dfrac{1}{640}=\dfrac{a}{10^b}$, where $a$ and $b$ are integers.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. Ackbach
2. kaliprasad
3. castor28

Solution from Ackbach:

Spoiler

We have
$$a=\frac{10^b}{640}=\frac{10^{b-1}}{64}=\frac{10\cdot 10^{b-2}}{64}=\frac{5\cdot 10^{b-2}}{32}
=\frac{5^2\cdot 10^{b-3}}{16}=\dots=5^6\cdot 10^{b-7}. $$
Hence, for $5^6\cdot 10^{b-7}$ to be an integer, we need $b\ge 7$; the smallest such $b$ is $7$, which makes $a=5^6$.