Let $$a$$ be an integer such that $$\lim_{x \to 7} \frac{18 - [1-x]}{[x-3a]}$$ exists, where $$[t]$$ is greatest integer $$\leq t$$. Then $$a$$ is equal to

We need to find the integer $$a$$ such that $$\displaystyle\lim_{x \to 7} \frac{18 - [1-x]}{[x-3a]}$$ exists, where $$[t]$$ denotes the greatest integer function (floor function).

As $$x \to 7^-$$, we have $$1 - x \to -6^+$$, so $$[1-x] = -6$$ and the numerator becomes $$18 - (-6) = 24$$. As $$x \to 7^+$$, $$1 - x \to -6^-$$, giving $$[1-x] = -7$$ and the numerator $$18 - (-7) = 25$$. Since the numerator has different left-hand and right-hand values (24 and 25), for the limit to exist, the denominator must also have a jump at $$x = 7$$ that compensates.

At $$x = 7$$ the expression in the denominator is $$x - 3a = 7 - 3a$$. If $$7 - 3a$$ is not an integer, then $$[x - 3a]$$ remains constant in a neighborhood of $$x = 7$$ and the limit would be $$\frac{24}{[7-3a]} \neq \frac{25}{[7-3a]}$$, so the limit would not exist. Therefore, $$7 - 3a$$ must be an integer (which is automatically satisfied since $$a$$ is an integer).

When $$7 - 3a$$ is an integer, as $$x \to 7^-$$ we have $$x - 3a \to (7-3a)^-$$ so $$[x - 3a] = 7 - 3a - 1 = 6 - 3a$$, and as $$x \to 7^+$$ we have $$x - 3a \to (7-3a)^+$$ so $$[x - 3a] = 7 - 3a$$.

Equating the left-hand limit and the right-hand limit yields $$\frac{24}{6 - 3a} = \frac{25}{7 - 3a}$$. Cross-multiplying gives $$24(7 - 3a) = 25(6 - 3a)$$, which simplifies to $$168 - 72a = 150 - 75a$$ and then to $$18 = -3a$$, so $$a = -6$$.

Verification: With $$a = -6$$, the denominator values are $$6 - 3(-6) = 24$$ and $$7 - 3(-6) = 25$$. Hence LHL = $$\frac{24}{24} = 1$$ and RHL = $$\frac{25}{25} = 1$$. The limit exists and equals 1.

The correct answer is Option C: $$-6$$.