The diagonal of the square with side 100 cm can be expressed as $a\times\sqrt b cm$ and b is Square Free.

Then X=ab

Let Y be the number of trailing zeroes in 100!

Let the roots of the quadratic equation ${x}^{2} + 100 x + 819$ be p and q

Then ${p}^{2} + {q}^{2} = Z$

Let A be the number of trailing zeroes in (X+Y+Z)!

Find $X+Y+Z+A$

The answer is 10729.

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I will just give away the answers.... I encourage you to solve them on your own.

The diagonal of the square can be expressed as $100\times\sqrt 2$ . Hence, X=200

The number of trailing zeroes in 100! is 24

The roots are -91 and -9... Therefore, Z= 8362

X+Y+Z = 8586. Therefore, the number of trailing zeroes in 8586! is 2143. Hence, A= 2143.

Adding them all up, you get 10729.

Cheers! :)