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Difference between revisions of "1978 AHSME Problems/Problem 4"

(Created page with "== Problem 4 == If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</math> is equal to <math...")
(No difference)

Revision as of 17:02, 20 January 2020

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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