recurring recurring recurring.....

Algebra Level 3

You are given the following recurring decimals 0.0123456789..... 0.1234567890.... 0.2345678901... and so on up until 0.9012345678..... What is the sum of the 10 decimals?


The answer is 5.

Curtis Clement by Curtis Clement , 23, United Kingdom

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1 solution

Ryan Tamburrino
Dec 30, 2014

The 9th triangular number (1+2+3+...+9) is 45. Because each of the given decimals has an infinite number of digits and is periodic, each column will have a 5 and a carried 4, so each digit after the decimal will be 5+4 = 9. The leftmost digit will be the ten's digit of the 9th triangular number, which is 4. The sum is 4.999999999... = 5

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The solution I intended was very similar to that. I used the fact that the sum goes 4.5 +0.45+0.045+... = 4.5(1+0.1+0.2+...) = 4.5 1 − 0.1 \frac{4.5}{1-0.1} = 9 2 \frac{9}{2} × \times 10 9 \frac{10}{9} = 5

Curtis Clement - 6 years, 5 months ago

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