Get to 0

Probability Level pending

How many ways are there to get to 0 by using the numbers 1 and 2 once only and any number of mathematical operators, symbols etc. (except the floor function and the modular operator)?

If you think the answer is infinity, input your answer as -1.


The answer is -1.

Freddie Hand by Freddie Hand , 18, United Kingdom

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2 solutions

Freddie Hand
Feb 26, 2017

s i n ( ( ( 1 + 2 ) ! ) ! ) = s i n ( 6 ! ) = s i n ( 720 ) = 0 sin(((1+2)!)!)=sin(6!)=sin(720)=0

s i n ( 720 ! ) = 0 sin(720!)=0

etc.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Phillip Temple
Feb 25, 2017

n mod 1 = 0 (For any integer n, All integers must be divisible by 1)

2 mod 1 = 0

2! mod 1 = 2 mod 1 = 0

(2!)! mod 1 = (2)! mod 1 = 2 mod 1 = 0

... This process can be repeated infinitely since the factorial is an operator and can be applied as many times as necessary. Therefore, there exist an infinite number of solutions.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I had in mind something different actually, so I changed the question a bit (you can't use mod any more). Well spotted!

Freddie Hand - 4 years, 3 months ago

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Here's another method; are the list of your exclusions going to grow longer?

The Heavside Step Function of α (denoted H(α)) is 1 for all positive numbers, ½ for 0, and -1 for all negative numbers.

So: H(2) - 1 = 1 - 1 = 0 H(2!) - 1 = H(2) - 1 = 0 ... Hence infinite solutions :D

Phillip Temple - 4 years, 3 months ago

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