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Answer: $399$
Using algebra
Let the five consecutive positive integers be $x-2$, $x-1$, $x$, $x + 1$, $x + 2$. Their sum is $5x$, so $5x = 2005$, that is $x = 401$. The five numbers are therefore $399, 400, 401, 402, 403,$ so the smallest number is $399$.
Using approximation/averages
The numbers are close to $2005\div5 = 401$
$$401+401+401+401+401=2005\\
401+400+401+402+401=2005\\
399+400+401+402+403=2005$$ The smallest number is $399$
Five-to
Age 11 to 14
ShortChallenge Level 
Answer: $399$
Using algebra
Let the five consecutive positive integers be $x-2$, $x-1$, $x$, $x + 1$, $x + 2$. Their sum is $5x$, so $5x = 2005$, that is $x = 401$. The five numbers are therefore $399, 400, 401, 402, 403,$ so the smallest number is $399$.
Using approximation/averages
The numbers are close to $2005\div5 = 401$
$$401+401+401+401+401=2005\\
401+400+401+402+401=2005\\
399+400+401+402+403=2005$$ The smallest number is $399$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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