Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.
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?
Integers
Digits
Sets
Answer: is 184.
AIME I, 1990, Question 13
Elementary Number Theory by David Burton
here \(9^{4000}\) has 3816 digits more than 9,
or, 4000-3816=184
or, 184 numbers have 9 as their leftmost digits.
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