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Find the largest integer n for which 12^n evenly divides 20!*40!.

 Apr 1, 2021
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\(12=3\cdot2^2\)

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 \(\boxed{26}\)

 Apr 1, 2021

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