Question #50079
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.
m=1000 then find the maximum possible value of k.
Expert's answer
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
m×(m+1) ×(m+2) ×(m+3)×… …×(3m-1) ×m
or
1000×1001 ×1003 ×1004×… …×2998 ×2999×1000
or
x/3
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#340153 on Dec 2023
#340153 on Dec 2023