Question
Prove that one of any three consecutive positive integers must be divisible by 3.
✓
Answer
Consider that if a number n, q and r are positive integers. When n is divided by 3 the quotient is a and remainder r. So, by Euclid's dividsion algorithm,
$\text{n}=3\text{q}+\text{r}\ (0\leq\text{r}<3)\ \text{or r}=0, 1, 2, 3$
So, one of any three consecutive positive integers is divisible by 3.
$\text{n}=3\text{q}+\text{r}\ (0\leq\text{r}<3)\ \text{or r}=0, 1, 2, 3$
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At
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$n_1$
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Divisible by 3
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$n_2 = n_1 + 3$
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Divisible by 3
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$n_3 = n_3 + 4$
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Divisible by 3
|
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r =0
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3q + 0 = 3q
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Yes
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3q + 1
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No
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3q + 2
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No
|
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r = 1
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3q + 1
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No
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3q + 2
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No
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3q + 3
=3(q + 1)
=3m
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Yes
|
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r = 2
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3q + 2
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No
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3q + 3 = 3(q + 1)
= 3m
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Yes
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3q + 4
=3q + 3 + 1
= 3(q + 1) + 1
= 3m + 1
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No
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