# Math problem?

The sum of three consecutive positive integers is perfect cube. Find the least possible value of the smallest number, assuming that the middle number ends in at least 5 zeros.

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• Anonymous
5 years ago

You realize that the sum of 3 consecutive integers is just 3 X the middle one, right?

x-1 + x + x+1 = 3x

So if 3x is a cube, x must be 3^2, or 3^2 times a cube.

3x9 = 27 = 8+9+10 = 3^3

Now, you want the middle number to end in at least 5 zeroes. That means it has 2^5 and 5^5 as factors. For it to be 9 x N where N is a cube, it must have at least 1 more 2 and 5 as factors. So the smallest number is 9 x 2^6 x 5^6 = 9 million,

8 999 999+

9 000 000+

9 000 001=

27 000 000 = 300^3

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• The sum of the three integers will just be three times the middle integer. This has to end in at least 5 zeros so it must have factors of 2^5 and 5^5. Three times this number must have at least factors of 2^5 . 3 . 5^5. To make a cube out of it we need to take each factor up to a power that's a multiple of 3, so that will give us

2^6 . 3^3 . 5^6 = (2^2 . 3 . 5^2)^3 = 300^3

The middle number is 2^6 . 3^2 . 5^6 = 9 . 10^6 = 9,000,000.

So the least possible value of the smallest number is 8,999,999.

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• Just know that the cube of 3 is 27.... then the consecutive numbers are

8, 9, 10 .... here the sum is 27 OK!

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