Difference between revisions of "Mock AIME 2 Pre 2005 Problems/Problem 4"
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Note that <math>5^n</math> has the same number of digits as <math>5^{n-1}</math> if and only if <math>5^{n-1}</math> has a leading digit <math>1</math>. Therefore, there are <math>2004 - 1401 = 603</math> numbers with leading digit <math>1</math> among the set <math>\{5^1, 5^2, 5^3, \cdots 5^{2003}\}.</math> However, <math>5^0</math> also starts with <math>1</math>, so the answer is <math>603 + 1 = \boxed{604}</math>. | Note that <math>5^n</math> has the same number of digits as <math>5^{n-1}</math> if and only if <math>5^{n-1}</math> has a leading digit <math>1</math>. Therefore, there are <math>2004 - 1401 = 603</math> numbers with leading digit <math>1</math> among the set <math>\{5^1, 5^2, 5^3, \cdots 5^{2003}\}.</math> However, <math>5^0</math> also starts with <math>1</math>, so the answer is <math>603 + 1 = \boxed{604}</math>. | ||
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== See also == | == See also == | ||
Latest revision as of 17:16, 4 August 2019
Problem
Let 
. Given that 
has 
digits, how many elements of 
begin with the digit 
?
Solution
Note that 
has the same number of digits as 
if and only if has a leading digit . Therefore, there are 
numbers with leading digit among the set 
However, 
also starts with , so the answer is 
.
-MP8148
See also
| Mock AIME 2 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
