+505 \(12=3\cdot2^2\)
So we need to find the largest value of \(2^{2n}\)\(\)that divides 20!*40!, which is
\(\frac{\lfloor\frac{20}{2}\rfloor+\lfloor\frac{20}{4}\rfloor+\lfloor\frac{20}{8}\rfloor+\lfloor\frac{20}{16}\rfloor+\lfloor\frac{40}{2}\rfloor+\lfloor\frac{40}{4}\rfloor+\lfloor\frac{40}{8}\rfloor+\lfloor\frac{40}{16}\rfloor+\lfloor\frac{40}{32}\rfloor}{2}\\ =\frac{10+5+2+1+20+10+5+2+1}{2}\\=28\)
And now the largest value of \(3^n\)which divides 20!*40!, which is
\(\lfloor\frac{20}{3}\rfloor+\lfloor\frac{20}{9}\rfloor+\lfloor\frac{40}{3}\rfloor+\lfloor\frac{40}{9}\rfloor+\lfloor\frac{40}{27}\rfloor\\ =6+2+13+4+1\\=26\)
Since 28>26, the largest integer n for which 12^n evenly divides 20!*40! is \(\boxed{26}\)
