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Algebra Difficulty 2.0 Junior Find the answer

If 991+993+995+997+999=5000−N991+993+995+997+999=5000-N, then N=N=

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Solution

991+993+995+997+999=5000−N⇒(1000−9)+(1000−7)+(1000−5)+(1000−3)+(1000−1)=5000−N⇒5×1000−(1+3+5+7+9)=5000−N⇒5000−25=5000−N⇒N=25→E.\begin{align*} 991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\ &\Rightarrow 5000-25=5000-N \\ &\Rightarrow N=25\rightarrow \boxed{\text{E}}. \end{align*}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.