1999 AMC 8 Problems/Problem 1

Problem

$(6?3) + 4 - (2 - 1) = 5.$ To make this statement true, the question mark between the 6 and the 3 should be replaced by

$\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}$

Simplify the given expression: $(6?3)+4-(2-1)=5$

$(6?3)+4-1=5$

$(6?3)+3=5$

$(6?3)=2$

At this point, it becomes clear that it should be $\div, \boxed{A}$ .

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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