1992 AIME Problems/Problem 6
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.
![\[\begin{array}{c|c|c|c|c|c|c} & & & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\ 0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\ 0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \end{array}\]](http://latex.artofproblemsolving.com/4/b/7/4b71412199a5a43049e774b7a9c8a8ff8a1fd660.png)
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.
See also
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
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