# Digits and Integers | AIME I, 1990 | Question 13

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

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.

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