Difference between revisions of "2003 AMC 12A Problems/Problem 5"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #5]] and [[2003 AMC 10A Problems|2003 AMC 10A #11]]}} | ||
== Problem == | == Problem == | ||
− | The | + | The sum of the two 5-digit numbers <math>AMC10</math> and <math>AMC12</math> is <math>123422</math>. What is <math>A+M+C</math>? |
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math> | <math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math> | ||
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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 = | + | Therefore, <math>A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14} </math>. |
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2003|ab=A|num-b=10|num-a=12}} | |
− | + | {{AMC12 box|year=2003|ab=A|num-b=4|num-a=6}} | |
− | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 14:35, 30 July 2011
- 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, .
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 |