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Algebra Level 3

On the range of values from [ 1 , 1000 ] [1, 1000] , there exist a certain number of perfect squares . As you move from one perfect square to the next without skipping any, you will find that the distance between perfect squares increases. What is the average of all such distances in the given range of values?


The answer is 32.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

Otto Bretscher
May 30, 2016

Relevant wiki: Arithmetic Progressions

The distances form an arithmetic sequence running from 2 2 − 1 2 = 3 2^2-1^2=3 to 3 1 2 − 3 0 2 = 61 31^2-30^2=61 , so that the average is 3 + 61 2 = 32 \frac{3+61}{2}=\boxed{32}

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Nice observation! I just left the squares in and noticed that it was a telescoping series because you would get 2 2 − 1 2 + 3 2 − 2 2 + 4 2 − 3 2 + . . . + 3 1 2 − 3 0 2 30 = 3 1 2 − 1 30 = 32 \dfrac{2^2 - 1^2 + 3^2 - 2^2 + 4^2 - 3^2 + ... + 31^2 -30^2}{30 } = \dfrac{31^2 - 1}{30} = \boxed{32}

Hobart Pao - 5 years ago

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Yes, very nice! (+1)

Otto Bretscher - 5 years ago

Haha, that's what I thought at first, and was wondering where the arithmetic progression was.

Nice problem, but I don't think it's well suited in the AP chapter.

Calvin Lin Staff - 5 years ago

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