Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Order.

## Digits and order – AIME I, 1992

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. Find number of ascending positive integers are there.

- is 107
- is 502
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Digits

Order

## Check the Answer

But try the problem first…

Answer: is 502.

AIME I, 1992, Question 2

Elementary Number Theory by David Burton

## Try with Hints

First hint

There are nine digits that we use 1,2,3,4,5,6,7,8,9.

Second Hint

Here each digit may or may not be present.

\(\Rightarrow 2^{9}\)=512 potential ascending numbers, one for subset of {1,2,3,4,5,6,7,8,9}

Final Step

Subtracting empty set and single digit set

=512-10

=502.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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