Interesting problem, here is my solution:
Firstly, 2^n is an even number, therefore a!, b! and c! must be either: 3 even numbers, or: 1 even and 2 odd numbers.
Factorials are all even except 1!
Path 1: 3 even numbers
We can prove that a=2 (1 < a < 3) or b=2 or c=2, otherwise the sum will be 6x (common factor of 3!) and 2^n is never divisible by 3.
Path 1-1: a=2 and b=2 and c=2
2!+2!+2!=6 is not power of 2
Path 1-2: a=2 and b=2
4+c! (c>=3) = 2^n
=> c!/4+1=2^m (m=n-2)
We can prove that c<4 otherwise c!/4 is even and c!+1 is odd number (2^m cannot be odd).
The only candidate is 3
2!+2!+3!=10 is not power of 2.
Path 1-3:
a=2 and b>=3 and c>=3
=> 2+6x+6y=2^n (x=a!/3! and y=b!/3!)
=> 3(x+y)+1=2^m (m=n-1)
We can prove that x+y must be odd and therefore b=3 or c=3 (otherwise both are divisible by 4
To simplify: 2+6+c!=2^n
=> 8+c!=2^n
=> c!/8+1=2^m (m=n-3)
We can prove that c<6 otherwise c!/8 is even and c!/8+1 is odd.
Candidates are 3,4,5
2!+3!+3!=14
3!+3!+3!=18
4!+3!+3!=36
None of them is power of 2
Updated flawed part after @rycharde pointing out missing solutions:
2!+3!+3!=14, not a power of 2
2!+3!+4!=32 (2^5)
2!+3!+5!=128 (2^7)
Path 2: 1+1+c!=2^n (2 even numbers)
then c!/2+1=2^m (m=n-1), we can prove that c<4 otherwise c!/2 is even number and c!/2+1 is odd number.
Candidates are c=2, 3, both satisfies the equation:
1!+1!+2!=4 (2^2)
1!+1!+3!=8 (2^3)
Therefore, all possible solutions are:
1!+1!+2!=2^2
1!+1!+3!=2^3
2!+3!+4!=2^5
2!+3!+5!=2^7
Too complicated for me lol
Very good, and legit method, but you've missed some solutions!
Upvoted all the same!
Hi, @rycharde, thank you for the problem and response. I should have triple checked before answering, especially when it was done by tapping solution on a phone and right before bed :p
The path 1-3 is flawed:
The last bit should be
2!+3!+3!=14, not a power of 2
2!+3!+4!=32 (2^5)
2!+3!+5!=128 (2^7)
Now the solution should be completed.
Thank you again for the interesting problem, I really enjoyed my time solving it ^_^
That's cool! We know that a,b and c cannot all be 3 or more, because the sum would then always be a multiple of 3, but one of them can be greater than 3.