Ask Question
1 January, 11:23

Prove that one of every three consecutive positive integer is divisible by 3

+4
Answers (1)
  1. 1 January, 12:42
    0
    Hello:

    all n in N; n (n+1) (n+2) = 3a a in N or : ≡ 0 (mod 3)

    1) n ≡ 0 (mod 3) ... (1)

    n+1 ≡ 1 (mod 3) ... (2)

    n+2 ≡ 2 (mod 3) ... (3)

    by (1), (2), (3) : n (n+1) (n+2) ≡ 0*1*2 (mod 3) : ≡ 0 (mod 3)

    2) n ≡ 1 (mod 3) ... (1)

    n+1 ≡ 2 (mod 3) ... (2)

    n+2 ≡ 3 (mod 3) ... (3)

    by (1), (2), (3) : n (n+1) (n+2) ≡ 1*2 * 3 (mod 3) : ≡ 0 (mod 3), 6≡ 0 (mod)

    3) n ≡ 2 (mod 3) ... (1)

    n+1 ≡ 3 (mod 3) ... (2)

    n+2 ≡ 4 (mod 3) ... (3)

    by (1), (2), (3) : n (n+1) (n+2) ≡ 2*3 * 4 (mod 3) : ≡ 0 (mod 3), 24≡ 0 (mod3)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Prove that one of every three consecutive positive integer is divisible by 3 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers