Difference between revisions of "2003 AMC 12A Problems/Problem 5"

Line 1: Line 1:
 +
{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #5]] and [[2003 AMC 10A Problems|2003 AMC 10A #11]]}}
 
== Problem ==
 
== Problem ==
The sume of the two 5-digit numbers <math>AMC10</math> and <math>AMC12</math> is <math>123422</math>. What is <math>A+M+C</math>?  
+
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>
 
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math>
Line 17: Line 18:
 
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>.  
 
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>.  
+
Therefore, <math>A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14} </math>.  
  
 
== See Also ==
 
== See Also ==
*[[2003 AMC 12A Problems]]
+
{{AMC10 box|year=2003|ab=A|num-b=10|num-a=12}}
*[[2003 AMC 12A/Problem 4|Previous Problem]]
+
{{AMC12 box|year=2003|ab=A|num-b=4|num-a=6}}
*[[2003 AMC 12A/Problem 6|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 13:35, 30 July 2011

The following problem is from both the 2003 AMC 12A #5 and 2003 AMC 10A #11, so both problems redirect to this page.

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, $A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions