Difference between revisions of "2006 Alabama ARML TST Problems/Problem 11"
(New page: ==Problem== The integer <math>5^{2006}</math> has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers <math>0\leq k \leq 2005</math> does <math>5^k</math> b...) |
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==Solution== | ==Solution== | ||
| − | Now either <math>5^k</math> starts with 1, or <math>5^{k+1}</math> has one more digit than <math>5^k</math>. From <math>5^0</math> to <math>5^ | + | Now either <math>5^k</math> starts with 1, or <math>5^{k+1}</math> has one more digit than <math>5^k</math>. From <math>5^0</math> to <math>5^{2005}</math>, we have 1401 changes, so those must not begin with the digit 1. <math>2006-1401=\boxed{605}</math> |
==See also== | ==See also== | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 11:13, 29 September 2008
Problem
The integer 
has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers 
does 
begin with the digit 1?
Solution
Now either starts with 1, or 
has one more digit than . From 
to 
, we have 1401 changes, so those must not begin with the digit 1. 
