1992 AIME Problem 2

Attempt Problem 2 of the 1992 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AIME solutions, or check the answer key.

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2.

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?

Answer: 502
Concepts:subsetsdigitscombinations
Difficulty rating: 1700
Small Hint:

Once a set of nonzero digits is chosen, their order is forced

Big Hint:

Exclude subsets of sizes 00 and 11 from the subsets of {1,2,,9}\{1,2,\ldots,9\}

Solution:

The digit 00 cannot occur, because it would have to be the first digit and leading zeroes are not part of a decimal representation. Every subset of at least two digits from {1,,9}\{1,\ldots,9\} gives exactly one ascending integer when written in increasing order. Hence the number is 29(90)(91)=51219=502.\begin{aligned}2^9-\binom90-\binom91&=512-1-9\\&=502.\end{aligned}

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