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Difference between revisions of "1975 AHSME Problems/Problem 6"

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Revision as of 12:21, 15 December 2016

The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$


Solution

Solution by e_power_pi_times_i


When the $n$th odd positive integer is subtracted from the $n$th even positive integer, the result is $1$. Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{\textbf{(E) } 80}$.

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