# abc = a! + b! + c!

#### Albert

##### Well-known member
A=abc=a!+b!+c!

here A is a 3-digit number

find A

#### eddybob123

##### Active member
Re: abc=a!+b!+c!

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

#### Albert

##### Well-known member
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

##### Well-known member
Re: abc=a!+b!+c!

145 = 1! + 4! + 5!

reason

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

##### Well-known member
MHB Math Helper
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
.

#### Albert

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

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

#### mathbalarka

##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!

Yes! nice, Zaid!

These are called factorions base 10. See, A014080.

Balarka
.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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.

#### Albert

##### Well-known member
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

##### Well-known member
MHB Math Helper
Re: abc=a!+b!+c!

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