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

 
(added problem and solution)
Line 5: Line 5:
  
 
== Solution ==
 
== 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 ==
 
== 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]]

Revision as of 19:13, 4 November 2006

Problem

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

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

Solution

$AMC10+AMC12=123422$

$AMC00+AMC00=123400$

$AMC+AMC=1234$

$2\cdot AMC=1234$

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

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

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

See Also

Invalid username
Login to AoPS