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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 has 3817 digits and that its first (leftmost) digit is 9, how many elements of have 9 as their leftmost digit?

Solution

Whenever you multiply a number by , the number will have an additional digit over the previous digit, with the exception when the new number starts with a , when the number of digits remain the same. Since has 3816 digits more than , exactly numbers have 9 as their leftmost digits.

See also

1990 AIME (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
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