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

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== Solution ==
 
== Solution ==
When a number is multiplied by <math>9</math>, the number will gain a digit, except when the new number starts with a 9.  
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When a number is multiplied by <math>9</math>, it gains a digit unless the new number starts with a 9.  
Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - (3817 - 1) = 184</math> numbers have 9 as their leftmost digits.
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Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, <math>4000 - 3816 = 184</math> numbers have 9 as their leftmost digits.
  
 
== See also ==
 
== See also ==

Revision as of 19:37, 15 March 2017

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

When a number is multiplied by $9$, it gains a digit unless the new number starts with a 9.

Since $9^{4000}$ has 3816 digits more than $9^1$, $4000 - 3816 = 184$ 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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