Difference between revisions of "2003 AMC 12A Problems/Problem 5"
(→Solution 3) |
|||
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>. | 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== | ==Video Solution== |
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.
Contents
[hide]Problem
The sum of the two 5-digit numbers and is . What is ?
Solution
Since , , and are digits, , , .
Therefore, .
Solution 2
We know that is more than . We set up and . We have . Solving for , we get . Therefore, the sum .
Solution 3
Consider the place values of the digits of and .
When we add and , must result in a units digit of , meaning is either or . Since is odd, this means a ten was carried over to the next place value from , and thus (as and the ten is carried over). Now, we know 3 is the units digit of , so is either or . Again, we must look at the digit before , or . is even, so must be less than , or else the ten would be carried over. Ergo, is . Nothing is carried over, so we have , and . Therefore, the sum of , , and is .
Solution 4
Intuitively, adding and is the same as adding to itself. We can divide by to get . So, = , = 1, and = . Adding all those up yields ~sdmd
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.