2006 Alabama ARML TST Problems/Problem 11

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. $2006-1401=\boxed{605}$

Page

Toolbox

Search

2006 Alabama ARML TST Problems/Problem 11

Problem

Solution

See also