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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

## Digits and Integers – AIME I, 1990

Let T={\(9^{k}\): k is an integer, \(0 \leq k \leq 4000\)} given that \(9^{4000}\) has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

- is 107
- is 184
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Digits

Sets

## Check the Answer

But try the problem first…

Answer: is 184.

Source

Suggested Reading

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

## Try with Hints

First hint

here \(9^{4000}\) has 3816 digits more than 9,

Second Hint

or, 4000-3816=184

Final Step

or, 184 numbers have 9 as their leftmost digits.

## 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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