Difference between revisions of "1990 AIME Problems/Problem 13"
(solution) |
m (→See also: c) |
||
Line 7: | Line 7: | ||
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=12|num-a=14}} | {{AIME box|year=1990|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 21:10, 24 November 2007
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
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 (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 |