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

Latest revision as of 12:00, 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)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

@@ Line 13: / Line 13: @@
 <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}}