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

Latest revision as of 06:35, 5 November 2024

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

Solution 4

Intuitively, adding $AMC10$ and $AMC12$ is the same as adding $AMC11$ to itself. We can divide $123422$ by $2$ to get $61711$ . So, $A$ = $6$ , $M$ = 1, and $C$ = $7$ . Adding all those up yields $6 + 1 + 7 = \boxed{\mathrm{(E)}\ 14}$ ~sdmd

Video Solution

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 28: / Line 28: @@
 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>.
+==Solution 4==
+Intuitively, adding <math>AMC10</math> and <math>AMC12</math> is the same as adding <math>AMC11</math> to itself. We can divide <math>123422</math> by <math>2</math> to get <math>61711</math>. So,
+<math>A</math> = <math>6</math>, <math>M</math> = 1, and <math>C</math> = <math>7</math>. Adding all those up yields <math>6 + 1 + 7 = \boxed{\mathrm{(E)}\ 14}</math> ~sdmd
 ==Video Solution==

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