Difference between revisions of "1990 AIME Problems/Problem 13"

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== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=12|num-a=14}}
 
{{AIME box|year=1990|num-b=12|num-a=14}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 21:10, 24 November 2007

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.

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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