Five-to

Age 11 to 14

ShortChallenge Level Yellow star

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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