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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 <math>24</math> 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= \boxed{\text{(E)14}</math>.
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We know that <math>AMC12</math> is <math>2</math> 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=61710</math>. Therefore, the sum <math>A+M+C= \boxed{\mathrm{(E)}\ 14}</math>.
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== Solution 3 ==
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Consider the place values of the digits of <math>AMC10</math> and <math>AMC12</math>.
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When we add <math>AMC10</math> and <math>AMC12</math>, <math>C + C</math> must result in a units digit of <math>4</math>, meaning <math>C</math> is either <math>2</math> or <math>7</math>. Since <math>M</math> is odd, this means a ten was carried over to the next place value from <math>C + C</math>, and thus <math>C = 7</math> (as <math>7 + 7 = 14</math> and the ten is carried over). Now, we know 3 is the units digit of <math>M + M + 1</math>, so <math>M</math> is either <math>1</math> or <math>6</math>. Again, we must look at the digit before <math>M</math>, or <math>A</math>. <math>A</math> is even, so <math>M</math> must be less than <math>5</math>, or else the ten would be carried over. Ergo, <math>M</math> is <math>1</math>. Nothing is carried over, so we have <math>A + A = 12</math>, and <math>A = 6</math>. Therefore, the sum of <math>A</math>, <math>M</math>, and <math>C</math> is <math>6 + 1 + 7 = \boxed{\mathrm{(E)}\ 14}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=ZetE-VohlFQ  ~David
  
 
== See Also ==
 
== See Also ==

Latest revision as of 21:56, 19 July 2023

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 $2$ more than $AMC10$. We set up $AMC10=x$ and $AMC12=x+2$. We have $x+x+2=123422$. Solving for $x$, we get $x=61710$. Therefore, the sum $A+M+C= \boxed{\mathrm{(E)}\ 14}$.

Solution 3

Consider the place values of the digits of $AMC10$ and $AMC12$.

When we add $AMC10$ and $AMC12$, $C + C$ must result in a units digit of $4$, meaning $C$ is either $2$ or $7$. Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$, and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$, so $M$ is either $1$ or $6$. Again, we must look at the digit before $M$, or $A$. $A$ is even, so $M$ must be less than $5$, or else the ten would be carried over. Ergo, $M$ is $1$. Nothing is carried over, so we have $A + A = 12$, and $A = 6$. Therefore, the sum of $A$, $M$, and $C$ is $6 + 1 + 7 = \boxed{\mathrm{(E)}\ 14}$.

Video Solution

https://www.youtube.com/watch?v=ZetE-VohlFQ ~David

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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