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If I have a number like $a$ written in decimal expansion form, and it never ends, when will multiplying it by an integer alter this property? (I mean the result terminates at a digit.)

To be more precise,

$$a\text{ is repeating}\ \wedge\ P(c) \implies ca \text{ terminates}$$

What is $P(c)$?

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2 Answers 2

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This can indeed happen, e.g. when $a = \frac17$ and $c = 7$. It will become a terminating decimal precisely when $c$ contains all factors of the denominator of $a$, except 2's and 5's.

To be more precise, if $a = \frac{x}{2^m 5^n y}$ with $x$ and $y$ relatively prime, and $y$ not divisible by 2 or 5, $c$ needs to be a multiple of $y$ in order for $ca$ to be a terminating decimal.

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  • $\begingroup$ See the edit I made. $\endgroup$
    – AHB
    Commented Jan 5, 2017 at 14:36
  • $\begingroup$ Yes, that's exactly what I wrote in my answer. I hope it's clear enough; if not, please let me know. $\endgroup$
    – Glorfindel
    Commented Jan 5, 2017 at 14:37
  • $\begingroup$ Aha. So the second paragraph is the in fact $P(c)$. yes? $\endgroup$
    – AHB
    Commented Jan 5, 2017 at 14:38
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – Glorfindel
    Commented Jan 5, 2017 at 14:38
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If your original number has a repeating decimal representation, then multiplying it by $99\cdots99$ (with as many nines as there are digits in the repeating period) will produce a terminating decimal.

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  • $\begingroup$ Is that the only possible way to terminate $a$? $\endgroup$
    – AHB
    Commented Jan 5, 2017 at 14:32
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    $\begingroup$ @AHB It's not the only way, but it is one way, and it is easy to find. Glorfindel's answer says how to find the other values of $c$ that will work. $\endgroup$
    – David K
    Commented Jan 5, 2017 at 14:34

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