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Albert
Well-known member
- Jan 25, 2013
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A=abc=a!+b!+c!
here A is a 3-digit number
find A
here A is a 3-digit number
find A
A=100a+10b+c=a!+b!+c!Are a, b, and c digits or are they positive integers?
Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next termThis is a very nice problem, Albert.
Actually, there are only finitely many numbers which are sum of the factorial of their own digits. (Prove why)
The number 145 is the penultimate term of the sequence of such numbers. If you don't mind, Albert, I give it as an exercise to find out the next term (via-computer approaches are welcome).
Balarka
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I'd prefer not telling that, that'd make things easier.Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term
The next one isI'd prefer not telling that, that'd make things easier.
A hint may suffice, for the sake of keeping this problem fair enough :
The next number is not too large.
Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.Hey Balarka , according to the link you provided there are only 4 numbers with this property , so is this proved or they are the only known integers .
They are proved, yes. If you look carefully a couple of posts back you'll see that I also asked for a proof of this fact.