Exponential prob: Define n# as n^{(n-1)^(n-2)^(n-3)^...^2^1}. Find last 2 digits of

[Steven G]

Define n# as n^{(n-1)^(n-2)^(n-3)^...^2^1}

What are the last two digits of the number 2018#?

 

[HallsofIvy]

What does that "#" mean? Did you intend to write 2018$?

 

[Steven G]

What does that "#" mean? Did you intend to write 2018$?

It was a typo which is now fixed. But to answer your question, $=#