Difference between revisions of "2006 Alabama ARML TST Problems/Problem 11"

(See also)
(Solution)
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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^{2005}</math>, we have 1401 changes, so those must not begin with the digit 1. <math>2006-1401=\boxed{605}</math>
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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>2005-1401=\boxed{604}</math>
  
 
==See also==
 
==See also==

Revision as of 22:08, 12 April 2012

Problem

The integer $5^{2006}$ has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers $0\leq k \leq 2005$ does $5^k$ begin with the digit 1?

Solution

Now either $5^k$ starts with 1, or $5^{k+1}$ has one more digit than $5^k$. From $5^0$ to $5^{2005}$, we have 1401 changes, so those must not begin with the digit 1. $2005-1401=\boxed{604}$

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 10
Followed by:
Problem 12
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