1993 UNCO Math Contest II Problems/Problem 2

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Determine the digit in the $623^{rd}$ place after the decimal point in the repeating decimal for: \[\frac{1}{9}+\frac{2}{99}+\frac{3}{999}.\]


Solution 1 =

Fist we add them together and divide. We get the repeating decimal .134316134316. We can see that it repeats every six decimal places, so we need to find the remainder when 623 is divided by six. When we divide, we see that the remainder is five, so we go in five decimal places and find 1, and we are done.

Solution 2

The sum of $1/9$ and $2/99$ is $13/99$, and the sum of of $13/99$ and $3/999$ (aka $1/333$) is $164/1221$. Doing long division, we get the decimal $0.\overline{134316}$, so the $623^{rd}$ place, we only need the $623\pod6=5^{th}$ digit, which is $\boxed{1}$

See also

1993 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions