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.
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 |
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