greatest X

From $n = \dfrac {61!+60!+59!} {3^x} = \dfrac {59!61^2} {3^x}$ , since $61$ is indivisible by $3$ , we have $3^x$ is the largest power of $3$ factor in $59!$ , .

Therefore,

$x = \left\lfloor \frac { 59 }{ 3 } \right\rfloor + \left\lfloor \frac { 59 }{ 3^2 } \right\rfloor + \left\lfloor \frac { 59 }{ 3^3} \right\rfloor$

$\quad = \left\lfloor \frac { 59 }{ 3 } \right\rfloor + \left\lfloor \frac { 59 }{ 9} \right\rfloor + \left\lfloor \frac { 59 }{ 27 } \right\rfloor$

$\quad = 19 + 6 + 2 = \boxed {27}$

{3}^x should divide 61!, 60! and 59!

so, first count the no. of trailing zeros in 59! as it's the least among three!

thus, 27.

so, it can have up to {3}^27 which can divide 59!.

thus, the answer is 27.