128 Zeroes

Relevant wiki: Trailing Number of Zeros

From this problem, we want to find $x$ such that $\left\lfloor \frac { x }{ 5 } \right\rfloor +\left\lfloor \frac { x }{ { 5 }^{ 2 } } \right\rfloor +\left\lfloor \frac { x }{ { 5 }^{ 3 } } \right\rfloor +\left\lfloor \frac { x }{ { 5 }^{ 4 } } \right\rfloor +\dots =128$ .

A good estimate is essential here.

If $x=300$ then $\left\lfloor \frac { 300 }{ 5 } \right\rfloor +\left\lfloor \frac { 300 }{ { 5 }^{ 2 } } \right\rfloor +\left\lfloor \frac { 300 }{ { 5 }^{ 3 } } \right\rfloor =60+12+2=74$ .

If $x=400$ then $\left\lfloor \frac { 400 }{ 5 } \right\rfloor +\left\lfloor \frac { 400 }{ { 5 }^{ 2 } } \right\rfloor +\left\lfloor \frac { 400 }{ { 5 }^{ 3 } } \right\rfloor =80+16+3=99$ .

If $x=500$ then $\left\lfloor \frac { 500 }{ 5 } \right\rfloor +\left\lfloor \frac { 500 }{ { 5 }^{ 2 } } \right\rfloor +\left\lfloor \frac { 500 }{ { 5 }^{ 3 } } \right\rfloor =100+20+4=124$ .

We need $4$ more factor $5$ 's, therefore $x=520$ .