2003 AMC 12A Problems/Problem 5
- 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 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 .
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.