Difference between revisions of "1990 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | When a number is multiplied by <math>9</math>, | + | 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 - | + | |
| + | 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 18:37, 15 March 2017
Problem
Let 
. Given that 
has 3817 digits and that its first (leftmost) digit is 9, how many elements of 
have 9 as their leftmost digit?
Solution
When a number is multiplied by 
, it gains a digit unless the new number starts with a 9.
Since 
has 3816 digits more than 
, 
numbers have 9 as their leftmost digits.
See also
| 1990 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
