Define n# as n(n-1)^(n-2)^(n-3)^...^2^1 What are the last two digits of the number 2018#?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,379 May 23, 2018 #1 Define n# as n(n-1)^(n-2)^(n-3)^...^2^1 What are the last two digits of the number 2018#? Last edited: May 24, 2018
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 May 24, 2018 #2 What does that "#" mean? Did you intend to write 2018$?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,379 May 24, 2018 #3 HallsofIvy said: What does that "#" mean? Did you intend to write 2018$? Click to expand... It was a typo which is now fixed. But to answer your question, $=#
HallsofIvy said: What does that "#" mean? Did you intend to write 2018$? Click to expand... It was a typo which is now fixed. But to answer your question, $=#