Answer to Question #50079 in Math for Quamrul

Question

x=m×(m+1) ×(m+2) ×(m+3) ×… …×(3m-1) ×3m. Here x is divisible by 3k, if
m=1000 then find the maximum possible value of k.

Answer 1

The maximum possible value of k is
m×(m+1) ×(m+2) ×(m+3)×… …×(3m-1) ×m
or
1000×1001 ×1003 ×1004×… …×2998 ×2999×1000
or
x/3