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2022 MMATHS Problems/Problem 1

Revision as of 19:58, 18 December 2022 by Arcticturn (talk | contribs) (→‎Problem)

Solution 1

We solve everything in terms of $a$. $b = 20 - a$ and $c = 2022 - a$. Therefore, $20-a + 2022-a = 22$. Solving for $a$, we get that $a$ = 1010. Since $c = 2022-a, c = 1012$. Since $b = 20-a, b = -990$. Computing $\frac {1010-(-990)}{1012-1010}$ gives us the answer of $\boxed {1000}$.

~Arcticturn

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