Random Problem

Revision as of 12:07, 30 January 2023 by Mathboy100 (talk | contribs) (Medium Problem)

Easy Problem

The sum\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\]can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution

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Medium Problem

Show that there exist no finite decimals $a = 0.\overline{a_1a_2a_3\ldots a_n}$ such that when its digits are rearranged to a different decimal $b = 0.\overline{a_{b_1}a_{b_2}a_{b_3}\ldots a_{b_n}}$, $a + b = 1$.

Solution

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