1992 AIME Problems/Problem 6
Contents
[hide]Problem
For how many pairs of consecutive integers in is no carrying required when the two integers are added?
Solution 1
For one such pair of consecutive integers, let the smaller integer be where
and
are digits from
through
We wish to count the ordered triples By casework, we consider all possible forms of the larger integer, as shown below.
Together, the answer is
~MRENTHUSIASM
Solution 2
Consider what carrying means: If carrying is needed to add two numbers with digits and
, then
or
or
. 6. Consider
.
has no carry if
. This gives
possible solutions.
With , there obviously must be a carry. Consider
.
have no carry. This gives
possible solutions. Considering
,
have no carry. Thus, the solution is
.
Solution 3
Consider the ordered pair where
and
are digits. We are trying to find all ordered pairs where
does not require carrying. For the addition to require no carrying,
, so
unless
ends in
, which we will address later. Clearly, if
, then adding
will require no carrying. We have
possibilities for the value of
,
for
, and
for
, giving a total of
, but we are not done yet.
We now have to consider the cases where , specifically when
. We can see that
, and
all work, giving a grand total of
ordered pairs.
Solution 4 (Complementary Counting)
Since there are numbers in the set, this means that there are
consecutive pairs of integers
in the set. For a pair to carry over, either the hundreds digits will carry over, the tens digits will carry over, or the ones digits will carry over.
For the hundreds digits, every consecutive pair in will have to carry over, except for the pair
when no carrying occurs, removing
pairs.
Now, consider the case when the tens digit is more than 5. Every pair of consecutive integers from will have to carry over except for the pair
, and since there are 5 of these cases less than
(namely the integers from
,
,
,
, and
) this removes
pairs from the list.
Now, consider the case where the ones digits carry over. In this case, every pair of consecutive integers from carries over, removing 4 pairs. Since there are 5 possible cases for the tens digit (
) and 5 possible cases for the hundreds digit, there are
more cases to be excluded.
In total, there are consecutive pairs.
~Soupboy0
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.