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 1402 changes (since 1 increases 1402 times to become 1403 after $5^{2005}$), so those must not begin with the digit 1. $2006-1402=\boxed{604}$

See also

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