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Show that of the numbers n , n + 2 , n + 4 , only one of them is divisible by 3 |
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Answer» he numbers n, (n+2), (n+4), where n is a positive integer.Now by dividing n by 3,LET Q be the quotient and r be the remainderso,n = 3q+rwhere 0 ≤ r < 3[By EDL- EUCLID's Division LEMMA]So, n = 3q+rRemainders (r) = 0, 1, 2Case 1, when r = 0n = 3q+0n+2 = 3q + 2n+4 = 3q + 4[Here n = 3q is only divisible by 3]Case 2, when r = 1n = 3q+1n+2 = 3q 2 + 1 = 3q + 3 n+4 = 3q+ 4 + 1 = 3q+5[Here (n+2) = (3q+3) is only divisible by 3]Case 3, when r = 2n = 3q+2n+2 = 3q+2 + 2 = 3q + 4 n+4 = 3q+ 4 + 2 = 3q + 6[Here n+4 = 3q+6 is only divisible by 3]HENCE,We conclude that from the numbers n, (n+2), (n+4), only one of them is divisible by 3 |
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