Question 246880
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You know the smallest already:  0 is neither positive nor negative, so the first positive integer is 1 and the sum of the three consecutive integers starting with 1 is 1 + 2 + 3 = 6.

Now we need to know the largest set. Let *[tex \Large x] be the largest of the three integers.  Then the next smaller consecutive integer must be *[tex \Large x\ -\ 1] and the one before that is *[tex \Large x\ -\ 2]

And

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ -\ 1\ +\ x\ -\ 2\ <\ 40]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ - 3\ <\ 40]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ < 43]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x < \frac{43}{3}]

Since 43 thirds is equal to 14 and one-third, the largest integer less than 43 thirds must be 14.

So, the largest set of three that sum to less than 40 is 12 + 13 + 14 = 39.

Check:  13 + 14 + 15 = 42 > 40, no good.

So, your first set is 1, 2, 3.  The second set is 2, 3, 4.  And so on until you get to 12, 13, 14.  You can fill in the missing sets.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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