Seja ABC um número de três dígitos (A é o dígito das centenas, B é o dígito das dezenas e C é o dígito das unidades, todos diferentes de 0). Seja n! a função fatorial, isto é, a multiplicação de todos os números naturais de 1 até n. Encontre três dígitos A, B e C que satisfaçam ABC = A! + B! + C!.
[Let ABC be a three-digit number (A is the hundreds digit, B is the tens digit and C is the units digit, all non-zero. Let n! be the factorial function, i.e., the multiplication of all natural numbers from 1 until n. Find three digits A, B and C that satisfy ABC = A! + B! + C!.]
PCFilho
The number is 145. 1!=1, 4!=24, and 5!=120. 1 + 24 + 120 = 145.
ResponderExcluirIn other words: A = 1; B = 4; and C = 5.
ExcluirThat’s it, Jake!
ExcluirImportant note: this is the one and only solution.
ResponderExcluirA, B or C cannot be 7, 8 or 9, because 7! = 5040, and this has 4 digits.
A, B or C cannot be 6, because 6! = 720, and then A would have to be 7, 8 or 9, and this is impossible.
Now, if all digits were less than 5, the largest sum we could produce would be 4! + 4! + 4! = 72. As the sum has to have 3 digits, at least one of them has to be 5.
One can easily check that the triples 551, 552, 553, 554 and 555 don't work. So, only one digit is 5.
Now, the largest sum we can have is 5! + 4! + 4! = 168. Then, A = 1.
As one of the other digits is 5, we are now left with only eight possibilities: four of the format 1B5 and four of the format 15C.
1! + 1! + 5! = 122
1! + 2! + 5! = 123
1! + 3! + 5! = 127
1! + 4! + 5! = 145 (the solution)
1! + 5! + 1! = 122
1! + 5! + 2! = 123
1! + 5! + 3! = 123
1! + 5! + 4! = 145