MathDB
Substituting powers of 10

Source: 1978 AHSME Problem 4

May 27, 2014
AMC

Problem Statement

If a=1, b=10, c=100a = 1,~ b = 10, ~c = 100, and d=1000d = 1000, then (a+b+c−d)+(a+b−c+d)+(a−b+c+d)+(−a+b+c+d)(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d) is equal to
<spanclass=′latex−bold′>(A)</span>1111<spanclass=′latex−bold′>(B)</span>2222<spanclass=′latex−bold′>(C)</span>3333<spanclass=′latex−bold′>(D)</span>1212<spanclass=′latex−bold′>(E)</span>4242<span class='latex-bold'>(A) </span>1111\qquad<span class='latex-bold'>(B) </span>2222\qquad<span class='latex-bold'>(C) </span>3333\qquad<span class='latex-bold'>(D) </span>1212\qquad <span class='latex-bold'>(E) </span>4242
Back to ProblemsView on AoPS