Express $x=0.\overline{31}_5$ as a fraction in lowest terms

Question

Given $x=0.\overline{31}_5$, find the value of $x$, expressed as a fraction in lowest terms.

I tried to change it into base $10$, but I don't think it's possible with fractions. So please help I'll appreciate it. Also I'm in 7th grade (easy solutions please) and no copying other people's answer (I've seen it in other problems).

Answer 1

$$x=.\overline{31}_5$$ $$100_5x=31.\overline{31}_5$$ subtracting the top from the bottom we get $44_5x=31_5$ (comment if you dont understand why $100_5-1_5=44_5$)

Solving for $x$ we get $x=\frac{31}{44}$ as a base $5$ fraction. To simplify, we convert to base 10 $$x=\frac{16}{24}=\frac{2}{3}$$

This works like converting base-10 recurring fractions (practice -- it's a good tool to have).

Answer 2

The usual approach to converting repeating decimals to fractions works here. Multiply it by $5$. Then subtract $5x-x$ and the repeating part disappears. You are left with $4x=$ something. Now convert that something to base $10$ because you are probably expected to give a result in base $10$.

Answer 3

Updated solution:

$0.31_5 = \frac{3}{5} + \frac{1}{25} = \frac{16}{25}$.

$0.\overline{31}_5 = 0.31_5 + 0.0031_5 + 0.000031_5 + \cdots$. Now using the formula for an infinite geometric series:

$$0.\overline{31}_5 = \frac{0.31_5}{1 - 0.01_5} = \frac{16/25}{1-1/25} = \frac{16}{24} = \frac{2}{3}.$$

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Old solution:

$x = 0.3_5 + 0.0\overline1_5$.

Now $0.\overline 4_5 = 1$. $0.0\overline1 \cdot 10_5 \cdot 4_5 = 1$, so $0.0\overline1 = \frac{1}{5 \cdot 4} = \frac{1}{20}$

Using the fact that $0.3_5 = \frac{3}{5}$, $x = \frac{3}{5} + \frac{1}{20} = \frac{13}{20}$.

Note: $0.\overline 4_5 = 1$ is the equivalent of $0.\overline{99} =1$. To prove this you can break it into $0.4_5 + 0.04_5 + 0.004_5 + \cdots$, which is an infinite geometric series and evaluates to $\frac{0.4_5}{1 - 0.1_5} = \frac{4/5}{1 - 1/5} = 1$.

Answer 4

An alternative approach is to use the geometric series formula; that is, for $|a|<1$ you have $$\sum_{n=1}^\infty a^n= \frac a{1-a}.$$

Now, $$.\overline{31}_5=3\cdot 5^{-1}+1\cdot 5^{-2}+3\cdot 5^{-3}+1\cdot 5^{-4}+\cdots=$$ $$=3\cdot 5\cdot 5^{-2}+1\cdot 5^{-2}+3\cdot 5\cdot 5^{-4}+1\cdot 5^{-4}+\cdots=$$ $$=16\cdot 5^{-2}+16\cdot 5^{-4}+\cdots=16\left(\frac1{25}+\left(\frac1{25}\right)^2+\cdots\right)=$$ $$=16\sum_{n=1}^\infty \left(\frac1{25}\right)^n= 16\cdot\frac {\frac1{25}}{1-\frac1{25}}=\frac23.$$