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 = 14 \ | + | Therefore, <math>A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14} </math>. |
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | We know that <math>AMC12</math> is <math>2</math> 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=61710</math>. Therefore, the sum <math>A+M+C= \boxed{\mathrm{(E)}\ 14}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Consider the place values of the digits of <math>AMC10</math> and <math>AMC12</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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=ZetE-VohlFQ ~David | ||
== 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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:56, 19 July 2023
- 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
.
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.