- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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

here A is a 3-digit number

find A

here A is a 3-digit number

find A

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

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

here A is a 3-digit number

find A

here A is a 3-digit number

find A

- Aug 18, 2013

- 76

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

- Thread starter
- #3

- Jan 25, 2013

- 1,225

A=100a+10b+c=a!+b!+c!Are a, b, and c digits or are they positive integers?

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

and a$\neq 0$

find A

- Mar 31, 2013

- 1,345

145 = 1! + 4! + 5!

reason

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

- Mar 22, 2013

- 573

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

.

- Thread starter
- #6

- Jan 25, 2013

- 1,225

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

.

- Mar 22, 2013

- 573

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

A hint may suffice, for the sake of keeping this problem fair enough :

The next number is not too large.

- Jan 17, 2013

- 1,667

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.

\(\displaystyle 40585 = 4!+0!+5!+8!+5!\)

- Mar 22, 2013

- 573

- Jan 17, 2013

- 1,667

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 .

- Mar 22, 2013

- 573

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 .

- Thread starter
- #12

- Jan 25, 2013

- 1,225

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

- Mar 22, 2013

- 573

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.