Prove that one of any three consecutive positive integers must be divisible by 3.
To do:
We have to prove that one of any three consecutive positive integers must be divisible by 3.
Solution:
Let $a = n, b = n + 1$ and $c =n + 2$
Order triplet is $(a, b, c) = (n, n + 1, n + 2)$
where, $n$ is any positive integer
At $n = 1$
$(a, b, c) = (1, 1 + 1, 1 + 2)$
$= (1, 2, 3)$
At $n = 2$
$(a, b, c) = (2, 2 + 1, 2 + 2)$
$= (2, 3, 4)$
At $n = 3$
$(a, b,c) = (3, 3 + 1, 3 + 2)$
$= (3,4,5)$
At $n =4$
$(a,b, c) =(4, 4 + 1, 4 +2)$
$= (4, 5, 6)$
At $n = 5$
$(a, b,c) = (5, 5 + 1, 5 +2)$
$= (5,6,7)$
At $n = 6$
$(a,b, c) = (6, 6 + 1, 6 + 2)$
$= (6,7,8)$
At $n = 7$
$(a, b,c) = (7, 7 +1, 7 + 2)$
$= (7,8,9)$
At $n = 8$
$(a, b,c) = (8,8+ 1, 8+ 2)$
$= (8,9,10)$
We observe that each triplet consists of one and only one number which is a multiple of 3.
Hence, one of any three consecutive positive integers must be divisible by 3.
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