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

m
(solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 digits and that its first (leftmost) digit is 9, how many elements of <math>T_{}^{}</math> have 9 as their leftmost digit?
+
Let <math>T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}</math>. Given that <math>9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost digit?
  
 
== Solution ==
 
== Solution ==
{{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.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=12|num-a=14}}
 
{{AIME box|year=1990|num-b=12|num-a=14}}

Revision as of 12:16, 3 March 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
Invalid username
Login to AoPS