We can rewrite the fraction as

[math]\frac{9s^2+3s+1}{27} - \frac{60\cdot27 + 26}{27(3s-1)}[/math]

For the result to be an integer, any fractional part of the left hand side must be cancelled out by the right hand side.

Let's first note that [math]9s^2 + 3s + 1 \equiv 0 \pmod{27}[/math] has no solutions. (You can get this by exhaustive search, or by completing the square, or even the quadratic formula, or simply observing that the quadratic is always of the form [math]3n+1[/math].) So the left hand side is never an integer. But it could have fractional part with numerator 1, 4, 7, 10, 13, 16, 19, 22, or 25.

We also can tell that s can't be too large, because then the fraction on the right hand side would be less than 1/27. More precisely

[math]\frac{60 \cdot 27 + 26}{27(3s-1)} \geq \frac{1}{27}[/math]


[math]s \leq 549[/math]

So at this point we can stop thinking and apply brute force, and verify that [math]s = 0[/math] and [math]s=549[/math] are the only solutions in nonnegative integers (though [math]s = 1[/math] gives another integral result.)

OK, now that we already know the answer, how can we avoid checking all 500-some cases? :)

Well, note that the right-hand fraction had better not include any factors on the bottom other than 27--- i.e., fractions of the form x/54, x/5, etc. won't work. Note that the left-hand side never reduces to lower terms (consult the list above.) That is, when does

[math]\displaystyle \frac{1646}{27(3s-1)} = \frac{x}{27}[/math]

in lowest terms? Well, that implies [math]x \cdot (3s - 1) = 1646[/math]. But looking at the prime factorization, [math]1646 = 2 \cdot 823[/math], so that severely constrains the possibilities:

[math]3s - 1 = 1[/math]

[math]3s - 1 = 2[/math]

[math]3s - 1 = 823[/math]

[math]3s - 1 = 1646[/math]

These give solutions [math]s = 2/3[/math], [math]s = 1[/math], [math]s = 824/3[/math], and [math]s = 549[/math], confirming our "brute force" results.
Written 18 Feb • View Upvotes
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