Question

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

Answer

Answer: 156

Concepts:digits casework whole number operations

Difficulty rating: 1780

Small Hint:

Write the smaller integer as $1abc$ and separate cases by the number of trailing $9$ s

Big Hint:

A digit that is unchanged must be at most $4$ , and the digit increased by $1$ must also pair without a carry

Solution:

Write the smaller number as $1abc.$ If $c\neq9,$ then $c\leq4$ and the unchanged digits $a$ and $b$ are each at most $4,$ giving $5^3=125$ pairs. If $c=9$ but $b\neq9,$ then $a\leq4$ and $b\leq4,$ giving $25$ pairs. If $b=c=9$ but $a\neq9,$ there are $5$ choices for $a.$ Finally, $1999+2000$ also needs no carry. The total is $125+25+5+1=156.$

1992 AIME Problem 6

Answer: 156

Difficulty rating: 1780

Small Hint:

Big Hint:

Solution:

Problem 6 in Other Years