2003 AMC 10A Problems/Problem 11: Difference between revisions

Latest revision as of 13:35, 30 July 2011

Redirect to:

@@ Line 1: / Line 1: @@
-== Problem ==
+#REDIRECT[[2003 AMC 12A Problems/Problem 5]]
-The sum of the two 5-digit numbers <math>AMC10</math> and <math>AMC12</math> is <math>123422</math>. What is <math>A+M+C</math>?
-<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math>
-== Solution ==
-<math>AMC10+AMC12=123422</math>
-<math>AMC00+AMC00=123400</math>
-<math>AMC+AMC=1234</math>
-<math>2\cdot AMC=1234</math>
-<math>AMC=\frac{1234}{2}=617</math>
-Since <math>A</math>, <math>M</math>, and <math>C</math> are digits, <math>A=6</math>, <math>M=1</math>, <math>C=7</math>.
-Therefore, <math>A+M+C = 6+1+7 = 14 \Rightarrow E </math>.
-== See Also ==
-{{AMC10 box|year=2003|ab=A|num-b=10|num-a=12}}
-[[Category:Introductory Algebra Problems]]