A Factorial Equation | Math Olympiads

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[Qermaq]

3:30 Should have mentioned that c>3 is impossible because 2 + c! = 2^n and all possible values of 2^n are congruent to 0 mod 4. Since all values of c! that are 4! or greater are divisible by 4, in this case c cannot exceed 3.

[bsmith]

Very nice problem. I solved it with a different approach.
I am going to separate the problem in cases depending on the value of n.

First case is n=1. Then a!+b!+c!=2. This has no solutions since the left side must be at least 3.
Second case is n=2. Then a!+b!+c!=4. 3!=6 is too big, so that leaves 2!=2 and 1!=1 as the only possible factorials. Then there is one solution 2!+1!+1!=2^2.
Now for n>=3. Take a!+b!+c!=2^n mod 8. Each of a!, b!, and c! can be 0(from 4! and larger), 1(from 1!), 2(from 2!), or 6(from 3!) mod 8 and 2^n mod 8 will always be 0 since n>=3. Working mod 8 we can have 0+0+0=0, 0+2+6=0, or 1+1+6=0.
Looking at 0+0+0=0 mod 8 then all the factorials must be 4! or bigger. But then the left side is a multiple of 3 as each factorial will have a common factor of 3. This contradicts the right side being a power of 2, so no new solutions from this case.
Then 0+2+6=0, mod 8. 2!=2 and 3!=6 are mandatory, then a!+8=2^n. Now if a>=6 then a! is a multiple of 16, call it 16x. Then 16x+8=8*(2x+1)=2^n. But then we have an odd factor on the left equaling a power of 2, thus a cannot be 6 or greater. So two specific values to check a=4 and a=5. 4!+3!+2!=32=2^5 and 5!+3!+2!=128=2^7. Both are solutions.
Finally 1+1+6=0, mod 8. All three factorials 1!=1, 1!=1 and 3!=6 are mandatory. 3!+1!+1!=8=2^3 checks out so is out last solution.
In summary the solutions for a!+b!+c!=2^n are 2!+1!+1!=2^2, 4!+3!+2!=2^5, 5!+3!+2!=128=2^7, and 3!+1!+1!=2^3.

[MisterPenguin]

{2, 0, 0}?
2+1+1=2^n if n = 2.

I may have misunderstood the solutions we wanted.

[l.w.paradis]

Cool how you showed these were all possible soutions.

[georgesbv]

wlog a<=b<=c means you have to put back the permutations after

[BlaqRaq]

I was “bolden” (cricket term) by this one.

[mrityunjaykumar]

why b can't be >3? you didn't clear it out... also if c>=6 in general.. then c!+2+b!=2^n ...now if n>=4 and c>=6...b!=14(mod 16) which is impossible you can check it... now if n<4 we don't have solutions since lhs>rhs...hence c in general is <6

[yoav]

Nice, i think you forgot the case a=2 and b>=4 this has no solutions because b and c>=4 so 2+b!+c!=2(mod8) and 2^n=0(mod8).

[eckhardfriauf]

Thrilling, inspiring, and entertaining! =)

[DrDeuteron]

is there any bitwise binary stuff you can do to solve this, since the RHS is 1[n-1 zeros]?

[KJIUYHN]

nice problem!
i solved it because i have done a similar problem before.

[conrad]

Are there more solutions of the scheme (1;1; c>3) ? .. Ah, in the end you discuss it

[goldfing]

Sybermath, I like the variety of equations, but often I am not sure which set of numbers the equation shall be solved. Is it over the integer. real or complex numbers? This strongly influences the solution set. Often you present difficult Looping equations which have a simple integer solution and ot seems that this type of equation is usually not analytically solvable, but can be solved only when it has integer solutions.

[conrad]

Regarding the case (2;2;c) I think your argumentation is not totally correct:
2^(n-2) - 1 is even for n=2 ... this path reveals no further solution as c! = 0 cannot be solved for integers.

[vrljk]

3:24 y r there no more solutions for this case?

[wtspman]

You forgot to test cases where a and b equal zero. 0! = 1.