solakis1
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find A,B,C such that:
ABC= A!+B!+C!
ABC= A!+B!+C!
The discussion focuses on finding integers A, B, and C such that the product ABC equals the sum of their factorials, represented as A! + B! + C!. It is established that A, B, and C cannot be 7, 8, or 9 due to the factorial values exceeding three digits, specifically noting that 7! equals 5040. Participants emphasize the need for explicit mention of the theorems, definitions, or axioms utilized in solving this factorial equation.
PREREQUISITESMathematicians, educators, and students interested in combinatorial mathematics and number theory, particularly those exploring factorial equations and integer solutions.
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?