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

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== Solution 2 ==
 
== Solution 2 ==
  
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We know that <math>AMC12</math> is 2 more than <math>AMC10</math>. We set up <math>AMC10=x</math> and <math>AMC12=x+2</math>. We have <math>x+x+2=123422</math>. Solving for <math>x</math>, we get <math>x=6170</math>. Therefore, the sum <math>A+M+C=14</math>.
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We know that <math>AMC12</math> is <math>24 more than </math>AMC10<math>. We set up </math>AMC10=x<math> and </math>AMC12=x+2<math>. We have </math>x+x+2=123422<math>. Solving for </math>x<math>, we get </math>x=6170<math>. Therefore, the sum </math>A+M+C=14$.
  
 
== See Also ==
 
== See Also ==

Revision as of 21:36, 7 December 2019

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

Solution 2

We know that $AMC12$ is $24 more than$AMC10$. We set up$AMC10=x$and$AMC12=x+2$. We have$x+x+2=123422$. Solving for$x$, we get$x=6170$. Therefore, the sum$A+M+C=14$.

See Also

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

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