find-n-in-ordre-to-have-7-divide-n-3-n-2-

Question Number 31043 by abdo imad last updated on 02/Mar/18

f i n d n i n o r d r e t o h a v e 7 d i v i d e n^{3} + n - 2

Commented by MJS last updated on 03/Mar/18

the sequence of remains of

n^{3} / 7 is ⟨ 1, 1, 6, 1, 6, 6, 0, \dots ⟩

n / 7 is ⟨ 1, 2, 3, 4, 5, 6, 0, \dots ⟩

their sum - 2 is

⟨ 0, 1, 0, 3, 2, 3, 5, \dots ⟩

so the solution is

⟨ 1, 3, 8, 10, 15, 17, \dots ⟩

= {1 + 7 k ∣ k \in N} \cup {3 + 7 k ∣ k \in N}

Commented by rahul 19 last updated on 02/Mar/18

1 .

Commented by MJS last updated on 03/Mar/18

k \in Z is possible too

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