solakis1
- 407
- 0
find A,B,C such that:
ABC= A!+B!+C!
ABC= A!+B!+C!
very good but can you explicity mention which are the theorem ,definitions or axiom used to solve the above?kaliprasad said:Because this is a 3 digit number each digit shall be < 7 as 7! is a 4 digit number that is 5040
Now no digit can be 6 because 6! = 720 so abc >700 so aother digit becomes 7
Largest digit has be 5 because if largest digit is 4 or more we have $ABC \le 72( 3 * 4!)$
So one digit is 5
Now 5!= 120
2 digits cannot be 5 because 5! + 5! = 240 and one dgit has to be 2 and it does not satisfy.
So we have 1 digit 5 and anothee digit A = 1.
B cannot be 5 because then largest C is 4 and sum <= 145 but b =5 makes abc > 150
So C is 5 and we have 100 + 10 * B + 5 = 1 + B! + 120 $ but tying different values B = 4 so number 145
solakis said:very good but can you explicity mention which are the theorem ,definitions or axiom used to solve the above?