Question

Prove that 1! +2!+3!..........+2007! Is not a perfect cube

Answer 1

Answer:

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n! is a multiple of 7 for n≥7. Hence the sum is ≡1!+2!+3!+4!+5!+6!≡5(mod7). Hence if your sum is a square a2, then a6≡53≡6(mod7), and if it is a cube a3, then a6≡52≡4(mod7). But a6≡1(mod7) by Fermat if 7∤a , and a6≡0(mod7) if 7∣a. Therefore, the sum is neither a square nor a cube.