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11/2 Maths Problem
- Bryan Carmichael
- Feb 11, 2021
- 1 min read
In celebration of no occasion whatsoever, I have taken it upon myself to offer my own puzzle.
In this four part question (tiered by difficulty), mathematically prove that:
a) 0! equals 1
b) adding one to the product of four consecutive numbers will always give you a perfect square
c) any number divided by 0 will give an undefined value
d) support part c with a graph
@TheProphet for parts 3 and 4, there is a much more elegant solution which doesn't require the use of inequalities. For part 2, the calculations are necessary to prove that k(k+1)(k+2)(k+3) + 1 makes the perfect square...
By definition, a! = a * (a-1)!, therefore when a=1, 1! = 1 * 0! = 1. Hence 0! must be 1.
Let the four consecutive numbers be k, k+1, k+2 and k+3, respectively. Expand k(k+1)(k+2)(k+3) + 1 and you'll see a perfect square (which I'm too lazy to calculate).
Proof by graph. Otherwise, it can just be understood as axiomatic.
As above. Can be proven using quadratic inequalities and asymptotes.