Difference between revisions of "2003 AMC 12A Problems/Problem 5"
A1337h4x0r (talk | contribs) (→Solution) |
|||
| Line 18: | Line 18: | ||
Since <math>A</math>, <math>M</math>, and <math>C</math> are digits, <math>A=6</math>, <math>M=1</math>, <math>C=7</math>. | Since <math>A</math>, <math>M</math>, and <math>C</math> are digits, <math>A=6</math>, <math>M=1</math>, <math>C=7</math>. | ||
| − | Therefore, <math>A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14} </math>. | + | Therefore, <math>A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14} </math>. |
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | 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>. | ||
== See Also == | == See Also == | ||
Revision as of 20:36, 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.
Contents
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 2 more than . We set up 
and 
. We have 
. Solving for 
, we get 
. Therefore, the sum 
.
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. 
