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Number Theory Level 4

Suppose we write the infinite decimal expansion for $\dfrac1n$ for any natural number $n> 1$ such that it is non-terminating. For example $\dfrac12$ can be expressed as $0.4\overline9$ as its infinite decimal expansion.

Denote $v_p (n)$ as the highest power of $p$ that divides $n$ . Determine the length of the non-periodic part of the infinite decimal expansion of $\dfrac1n$ .

Details and Assumptions

As an explicit example, $200 = 2^3 \times 5^2$ , so $v_2 (200) = 3$ and $v_5(200) = 2$ because $2^3 | 200 , 2^4 \not | \ 200$ and $5^2 | 200, 5^3 \not | \ 200$ .

\max(v_3(n), v_6(n))

\max(v_3(n), v_5(n))

\max(v_2(n), v_5(n))

\max(v_4(n), v_5(n))

1 solution

Andy Hayes
Sep 22, 2015

Let $n=2^a\times5^b\times m$ , where $\gcd(2,m)=1$ and $\gcd(5,m)=1$ .

Case 1: $a>b$

$\frac{1}{n}=\frac{1}{2^a\times5^b\times m}=\frac{1\times\frac{5^{a-b}}{m}}{2^a\times5^b\times m\times\frac{5^{a-b}}{m}}=\frac{\frac{5^{a-b}}{m}}{10^a}$

In this case, $a$ represents the length of the non-repeating part of the decimal expansion. When $\frac{5^{a-b}}{m}$ is written as a mixed number, the whole number part of that mixed number will represent the non-repeating digits in the decimal expansion.

As an explicit example, let $n=120$ . We can re-write this as $n=2^3\times5^1\times3$ . We can then re-write $\frac{1}{n}=\frac{\frac{25}{3}}{1000}=\frac{8\frac{1}{3}}{1000}$ . The whole number part of the numerator, $8$ , is the non-repeating part of the decimal expansion. Since the denominator is $10^3$ , this non-repeating part is 3 digits long: $008$ . We can see this is true from the decimal expansion of $\frac{1}{120}$ : $0.008\overline{3}$ .

Case 2: $a<b$

$\frac{1}{n}=\frac{1}{2^a\times5^b\times m}=\frac{1\times\frac{2^{b-a}}{m}}{2^a\times5^b\times m\times\frac{2^{b-a}}{m}}=\frac{\frac{2^{b-a}}{m}}{10^b}$

Similarly to Case 1, $b$ represents the length of the non-repeating part of the decimal expansion. When $\frac{2^{b-a}}{m}$ is written as a mixed number, the whole number part of that mixed number will represent the non-repeating digits in the decimal expansion.

As an explicit example, let $n=4375=2^0\times5^4\times7$ . We can re-write $\frac{1}{4375}=\frac{\frac{16}{7}}{10000}=\frac{2\frac{2}{7}}{10000}$ . The non-repeating part of the decimal is represented by the whole number part of the numerator, $2$ . The denominator is $10^4$ , so the non-repeating part is 4 digits long. We can see this is true from the decimal expansion of $\frac{1}{4375}$ : $0.0002\overline{285714}$ .

Case 3: $a=b$

$\frac{1}{n}=\frac{1}{2^a\times5^b\times m}=\frac{1}{10^a\times m}=\frac{1\times\frac{1}{m}}{10^a\times m\times\frac{1}{m}}=\frac{\frac{1}{m}}{10^a}$

In this case, the non-repeating part of the decimal expansion is composed entirely of $0$ s, and its length is $a$ .

It can be shown from these cases that the length of the non-repeating part of the decimal expansion of $\frac{1}{n}$ is $\max(a,b)$ .

Don't dare to try it!

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