Hi,

today I am going to write about “my own” work. Specifically the solution to the three 3s puzzle which I encountered in physics camp (where as a morning routine we solved some problems) and nobody (I guess) knew solution back then.

It goes like this: Using plus, minus, division, multiplication, square roots, exponents, brackets and factorials and three 3s, get the number 1 through 10. All except 10 are quite simple, the solutions go like this (there are more options):

(3+3)/3!=1

(3+3)/3=2

(3/3)*3=3

….

And so on. It is important to remember that while you can use “the square root” you can not use “cubic root” without using 3. Also you can not write 1/3 because you would use one which is not allowed. To explain how my solution works you have to understand the difference between double factorial and factorial used two times. Factorial is defined as follows:

n!=n*(n-1)*(n-2)*…*2*1

If you use the factorial again like this: (n!)! you will get the factorial of the new number, it will grow quickly:

3!=6

(3!)!=6!=720

On the other hand:

n!!=n*(n-2)*(n-4)*4*2 (depends if odd or even)

n!!!=n*(n-3)*(n-6)*4*1 (the last numbers depend on the first again)

This way the factorial is defined is extremely useful, because then you can use it to multiply your number with almost any lower number. This is where my expression comes from:

(((3!)!!)!^43)/((3!)!!!!+(3!)!!!!)=10

Now I need (3!)!!=48 to be multiplied by 5, so I added 43 factorials.

If you search for this riddle on the internet, you will get often solutions which need to use other functions, while some of them are nice: log_{√√3}(3) + 3! = 10. I still find my solution to be the best. Most of those you will find use something obscure like flooring or rounding of numbers which makes it extremely easy. My method only uses the fact that the factorial is defined like that when you stack it on itself, I don’t think that without this property, it is possible to solve the problem.

Dragallur

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