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Question

Prove that one of any three consecutive positive integers must be divisible by 3.


Solution

Consider a number $$n, q$$ and $$r$$ are positive integers. When $$n$$ is divided by 3 the quotient is $$q$$ and remainder $$r$$. So, by Euclid;s division algorithm,
$$n = 3q + r (0 \le r < 3)$$ or $$r = 0, 1, 2, 3$$
 At$$n_1$$ Divisible by 3 $$n_2 = n_1 + 1$$ Divisible by 3$$n_3 = n_1 + 1$$  Divisible by 3 
$$r=0$$ $$3q+0=3q$$  Yes$$3q+1$$ No $$3q+2$$ No 
 $$r=1$$$$3q+1$$  No $$3q+2$$ No  $$3q+3$$
$$=3(q+1)$$
$$=3m$$
 Yes
 $$r=2$$ $$3q+2$$ No  $$3q+3$$
$$=3(q+1)$$
$$=3m$$
 Yes $$3q+4$$
$$=3q+3+1$$
$$=3(q+1)+1$$
$$=3m+1$$
 No 
So, one of any three consecutive positive integers is divisible by 3.

Mathematics

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