1990 AIME Problems/Problem 13

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Problem

Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . 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?

Solution

Whenever you multiply a number by $9$ , the number will have an additional digit over the previous digit, with the exception when the new number starts with a $9$ , when the number of digits remain the same. Since $9^{4000}$ has 3816 digits more than $9^1$ , exactly $4000 - (3817 - 1) = 184$ numbers have 9 as their leftmost digits.

1990 AIME (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1990 AIME Problems/Problem 13

From AoPSWiki

Problem

Solution

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