Find A: 3-Digit Number from Abc = a + b + c

[Albert1]

A=abc=a!+b!+c!

here A is a 3-digit number

find A

 

[eddybob123]

Re: abc=a!+b!+c!

Are a, b, and c digits or are they positive integers?

 

[Albert1]

Re: abc=a!+b!+c!

Are a, b, and c digits or are they positive integers?

A=100a+10b+c=a!+b!+c!

a,b,c $ \subset$ { 0,1,2,3,4,5,6,7,8,9 }

and a$\neq 0$

find A

 

[kaliprasad]

Re: abc=a!+b!+c!

Spoiler

145 = 1! + 4! + 5!

reason

Spoiler

none of abc can be > 5 as 6! = 720 and 7! = 5040 > 1000

one of them that is b or c= 5 ( a cannot be 5 as 5! = 120 and 5! + 4! + 3! < 200)

so a = 1, b= 5, c = ? or a = 1, b = ? , c = 5 ( it has to be < 5)

if a = 1 , b = 5 we get 1 + 120 + c ! > 150 and < 160

so c! > 29 so there is no c

if a = 1, c = 5 we get 1 + 120 + b! = 105 + 10 b

so b = 4

 

[mathbalarka]

Re: abc=a!+b!+c!

This 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
.

 

[Albert1]

Re: abc=a!+b!+c!

This 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
.

Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term

 

[mathbalarka]

Re: abc=a!+b!+c!

Here A is 3- digit number , please tell me the numbers of digits you want me to find for the next term

I'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.

 

[alyafey22]

Re: abc=a!+b!+c!

I'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.

The next one is

Spoiler

$$40585 = 4!+0!+5!+8!+5!$$

 

[mathbalarka]

Re: abc=a!+b!+c!

Yes! nice, Zaid!

These are called factorions base 10. See, A014080.

Balarka
.

 

[alyafey22]

Re: abc=a!+b!+c!

Yes! nice, Zaid!

These are called factorions base 10. See, A014080.

Balarka
.

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 .

 

[mathbalarka]

Re: abc=a!+b!+c!

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.

 

[Albert1]

Re: abc=a!+b!+c!

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.

ans :1, 2, 145, 40585
I wrote a program (using Excel) and found no answer for 4 digits number
and the only five digits number is 40585
the first person proved this (if using computer not allowed) must be very smart :)

 

[mathbalarka]

Re: abc=a!+b!+c!

Proving finiteness of the sequence is not hard. Note that any n-digit factorion has an upper bound $$n 9!$$ and a lower one $$10^{(n-1)}$$. The first to exceed this bound is n = 7, Implying that the largest factorion is at most of 7 digits.