1978 AHSME Problems/Problem 4: Difference between revisions

Latest revision as of 11:00, 13 February 2021

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

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

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

1978 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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 <cmath>3(a+b+c+d)</cmath>
 This is equal to <math>3(1111) = \boxed{\textbf{(C) }3333}</math> ~awin
+==See Also==
+{{AHSME box|year=1978|num-b=3|num-a=5}}
+{{MAA Notice}}