An old math teacher of mine gave me this fun little problem: For each number n from 1 to 9, make 6 from three copies of n, using only basic-ish arithmetic and no other numbers. For example, 2+2+2 = 6. I will leave the rest as a puzzle for the reader.
Precise statement of restrictions (enumerating these may give you a bit of a hint): You're allowed to use these operators: + - * / √ !
I did this, and then I figured out that if you're allowed to use the floor function as well, you can fulfill this task for any n ≥ 1. If [x] denotes the floor of x, then I'll demonstrate with 56:
√56 is somewhat less than 8. √√56 is somewhat less than 3. √√√56 is somewhat less than 2. Then [√√√56] = 1. Thus, we can have: ([√√√56]+[√√√56]+[√√√56])! = 3! = 6.
The expression to find "number of times you'll have to take square root of n" is slightly interesting: we want to take k square roots and end up with something less than 2, so n^(1/2^k) < 2, or n < 2^(2^k), so k > log_2(log_2(n)).
This reminds me of Knuth's conjecture. http://www.perlmonks.org/?node_id=443037 (Starting with 3, apply some combination of factorial, square root, and floor functions. Knuth conjectured all natural numbers are achievable this way.)
Precise statement of restrictions (enumerating these may give you a bit of a hint): You're allowed to use these operators: + - * / √ !
I did this, and then I figured out that if you're allowed to use the floor function as well, you can fulfill this task for any n ≥ 1. If [x] denotes the floor of x, then I'll demonstrate with 56:
√56 is somewhat less than 8. √√56 is somewhat less than 3. √√√56 is somewhat less than 2. Then [√√√56] = 1. Thus, we can have: ([√√√56]+[√√√56]+[√√√56])! = 3! = 6.
The expression to find "number of times you'll have to take square root of n" is slightly interesting: we want to take k square roots and end up with something less than 2, so n^(1/2^k) < 2, or n < 2^(2^k), so k > log_2(log_2(n)).