2003 AMC 12A Problems/Problem 5: Difference between revisions

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.

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

2003 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #5]] and [[2003 AMC 10A Problems|2003 AMC 10A #11]]}}
 == 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>
@@ 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>.
-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 ==
-*[[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]]