Don't forget the carry over

Algebra Level 2

0. 2 2 2 2 × 0. 5 5 5 5 0. 1 2 3 4 \begin{array}{cccccccc} && & 0. & 2 & 2 & 2&2\ \ldots \\ \times && & 0. & 5 & 5 & 5&5\ \ldots \\ \hline && & 0. & 1& 2 &3&4\ \ldots \\ \end{array}

The above shows a long multiplication, where only the first 4 decimal places are shown.

As you can see, the first 4 digits after the decimal place of the final product is 1234.

But what is the first 8 digits after the decimal place of the final product?


The answer is 12345679.

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2 solutions

Naren Bhandari
Oct 1, 2018

Note that { 0.222 = 0.2 + 0.02 + 0.002 + = n = 1 2 1 0 n 0.555 = 0.5 + 0.05 + 0.005 + = n = 1 5 1 0 n \begin{cases} 0.222\cdots =0.2 + 0.02 +0.002+\cdots = \displaystyle \sum_{n=1}^{\infty}\dfrac{2}{10^n} \\ 0.555\cdots =0.5+ 0.05+0.005+\cdots = \displaystyle \sum_{n=1}^{\infty}\dfrac{5}{10^n} \end{cases} Thus above result gives us n = 1 2 1 0 n = 2 9 n = 1 5 1 0 n = 5 9 \sum_{n=1}^{\infty}\dfrac{2}{10^n} = \dfrac{2}{9}\qquad \sum_{n=1}^{\infty}\dfrac{5}{10^n}=\dfrac{5}{9} Making the answer n = 1 2 1 0 n n = 1 5 1 0 n = 10 81 = 0.123456790 \sum_{n=1}^{\infty}\dfrac{2}{10^n}\cdot \sum_{n=1}^{\infty}\dfrac{5}{10^n}= \dfrac{10}{81}=0.123456790\cdots Thus, our required answer is 12345679 12345679 .

We know 0.2222 = 2 × 0.1111 = 2 × 1 9 = 2 9 0.2222\dots=2 \times 0.1111\dots=2 \times \frac{1}{9}=\frac{2}{9} & 0.5555 = 5 × 0.1111 = 5 × 1 9 = 5 9 0.5555\dots=5 \times 0.1111\dots=5 \times \frac{1}{9}=\frac{5}{9}

So the answer is 2 9 × 5 9 = 10 81 0.12345679 \frac{2}{9} \times \frac{5}{9}=\frac{10}{81} \approx \boxed{\large{0.12345679}}

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