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

Revision as of 19:35, 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 $9$ , the number will gain a digit, except 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.

@@ Line 3: / Line 3: @@
 == Solution ==
-Whenever you multiply a number by <math>9</math>, the number will have an additional digit over the previous digit, with the exception when the new number starts with a <math>9</math>, when the number of digits remain the same. Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, exactly <math>4000 - (3817 - 1) = 184</math> numbers have 9 as their leftmost digits.
+When a number is multiplied by <math>9</math>, the number will gain a digit, except when the new number starts with a <math>9</math>, when the number of digits remain the same. Since <math>9^{4000}</math> has 3816 digits more than <math>9^1</math>, exactly <math>4000 - (3817 - 1) = 184</math> numbers have 9 as their leftmost digits.
 == See also ==

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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Difference between revisions of "1990 AIME Problems/Problem 13"

Revision as of 19:35, 15 March 2017

Problem

Solution

See also