Difference between revisions of "2023 AMC 10B Problems/Problem 18"

Revision as of 15:23, 15 November 2023

$I.$ Try $a=3,b=5 => c = 17\cdot15$ which makes $\textbf{I}$ false. At this point, we can rule out answer A,B,C.

$II.$ A => B or C. equiv. ~B AND ~C => ~A. Let a = 14, b=15 (statisfying ~B and ~C). => C = 2*210. which is ~A.

$II$ is true.

So the answer is E. $\boxed{\textbf{(E) } II \text{ and } III \text{only}.}$ ~Technodoggo

@@ Line 1: / Line 1: @@
-The answers will be uploaded soon
+== Solution ==
-- MRP
+<math>I.</math>  Try <math>a=3,b=5 => c = 17\cdot15</math> which makes <math>\textbf{I}</math> false.
+At this point, we can rule out answer A,B,C.
+<math>II.</math> A => B or C. equiv. ~B AND ~C => ~A.
+Let a = 14, b=15 (statisfying ~B and ~C). => C = 2*210. which is ~A.
+<math>II</math> is true.
+So the answer is E.
+<math>\boxed{\textbf{(E) } II \text{ and } III \text{only}.}</math>
+~Technodoggo