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 Squares and Cubes II (Posted on 2013-05-17)
5 consecutive positive integers are given such that the sum of the middle 3 numbers is a perfect square, and the sum of all 5 numbers is a perfect cube. What is the minimum possible value of the first integer?

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 3.0000 (1 votes)

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Let x be the middle of the five integers, and therefore also the middle of the three middle integers.

The three middle integers add to 3x and the five digits altogether add to 5x, since the middle digit of each is the average of each set.

3x is a square so we need to have 3 as a factor of x an odd number of times, and any other prime an even number of times.

5x is a cube so 5 needs to be a factor twice, or five times, etc. Let it be twice to satisfy the previous paragraph. Any other prime needs to be a appear as a factor a multiple of three times.

3*3*3*5*5 = 675 works, and is the smallest that does.

Since that's the middle of the five numbers, the five are:

673, 674, 675, 676, 677

the first of which is 673, the answer.

 Posted by Charlie on 2013-05-17 12:09:37

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