Difference between revisions of "2003 AMC 10A Problems/Problem 11"

Revision as of 19:13, 4 November 2006

The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$ ?

$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$AMC+AMC=1234$

$2\cdot AMC=1234$

$AMC=\frac{1234}{2}=617$

Since $A$ , $M$ , and $C$ are digits, $A=6$ , $M=1$ , $C=7$ .

Therefore, $A+M+C = 6+1+7 = 14 \Rightarrow E$ .

@@ Line 5: / Line 5: @@
 == Solution ==
-<math>M+M</math> is odd, so <math>C+C</math> must carry over, and therefore, <math>C=7</math>.
+<math>AMC10+AMC12=123422</math>
-<math>A+A=12</math>, so <math>A=6</math>.
+<math>AMC00+AMC00=123400</math>
-<math>M+M+1=3</math>, so <math>M=1</math>.
+<math>AMC+AMC=1234</math>
-<math>7+6+1=14\Longrightarrow\mathrm{(E)}</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 ==
-* [[2003 AMC 10A Problems]]
+*[[2003 AMC 10A Problems]]
-* [[2003 AMC 10A Problems/Problem 12 | Next Problem]]
+*[[2003 AMC 10A Problems/Problem 10|Previous Problem]]
-* [[2003 AMC 10A Problems/Problem 10 | Previous Problem]]
+*[[2003 AMC 10A Problems/Problem 12|Next Problem]]
+[[Category:Introductory Algebra Problems]]