+0

# number theory

0
80
1

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

Apr 1, 2021

#1
+420
0

$$12=3\cdot2^2$$

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