2006 Alabama ARML TST Problems/Problem 11

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

2006 Alabama ARML TST (Problems)
Preceded by: Problem 10	Followed by: Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2006 Alabama ARML TST Problems/Problem 11

From AoPSWiki

Problem

Solution

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