1992 AIME Problems/Problem 6

Problem

For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?

Solution

Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$ , then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$ . 6. Consider $c \in \{0, 1, 2, 3, 4\}$ . $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$ . This gives $5^3=125$ possible solutions.

With $c \in \{5, 6, 7, 8\}$ , there obviously must be a carry. Consider $c = 9$ . $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$ , $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$ .

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1992 AIME Problems/Problem 6

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