Reply To: Maths

[frazza238]

USB-C_is_supreme

@Astralica has exhausted all solutions as:
As 2^n is a logarithm, it will double for every increase of n.
1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120,
If a(or b)! > 4, the final digit will always be 0. 2^n cycles final digits in the order 2, 4, 8, 6. Therefore, both a and b cannot be greater than 4.
When a(or b) is greater than 4, the other must be 2, 4, 6. No factorial exists that is either 2, 6 or 24 less than any 2^n when n > 3.
This means that the resulting possibilities are:
1! + 1! = 2^1: 1 + 1 = 2 {1:1:1}
2! + 2! = 2^2: 2 + 2 = 4 {2:2:2}
3! + 2! = 2^3: 6 + 2 = 8 (and vice-versa) {3:2:3} {2:3:3}

So:
a, b <= 3 for valid solutions. If a, b > 4: will always end in 0 which does not occur in the 2^n series. For values a,b where a <= 3 and b>= 4:
a = 1: 2^n – 1 does not exist as a factorial as always odd. ( also followed for a=3).
a = 2: 2^n – 2 = 2(2^(n-1)-1) which is always singly even (only 1 factor of 2) therefore cannot be factorial >= 4

(This final bit i was assisted with. This is not entirely my own work.)