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1978 AHSME Problems/Problem 4

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Problem 4

If $a = 1,~ b = 10, ~c = 100$ , and $d = 1000$ , then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to

$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad \textbf{(E) }4242$

Solution 1

Adding all four of the equations up, we can see that it equals $3(a+b+c+d)$ . This is equal to $3(1111) = \boxed{\textbf{(C) }3333}$

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