Question

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Show that one and only one out of n, n+2 or n+4 is divisible by 3, where 'n' is any positive integer.​

Answer 1

$\: \: \: \huge {\colorbox{pink}{▶Question :-}}$

➥ Show that one and only one out of n, n + 2 and n + 4 is divisible by 3, where 'n' is any positive integer .

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$\bf {\colorbox{pink}{▶Step-by-step explanation :-}}$

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

∵ Thus any number is in the form of 3q , 3q+1 or 3q+2.

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➺ Case I: if n =3q

⇒n = 3q = 3(q) is divisible by 3,

⇒ n + 2 = 3q + 2 is not divisible by 3.

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

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➺ Case II: if n =3q + 1

⇒ n = 3q + 1 is not divisible by 3.

⇒ n + 2 = 3q + 1 + 2 = 3q + 3 =

3(q + 1) is divisible by 3.

⇒ n + 4 = 3q + 1 + 4 = 3q + 5 =

3(q + 1) + 2 is not divisible by 3.

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➺ Case III: if n = 3q + 2

⇒ n =3q + 2 is not divisible by 3.

⇒ n + 2 = 3q + 2 + 2 = 3q + 4 =

3(q + 1) + 1 is not divisible by 3.

⇒ n + 4 = 3q + 2 + 4 = 3q + 6 =

3(q + 2) is divisible by 3.

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Thus one and only one out of n , n+2, n+4 is divisible by 3.

$\large \sf{Hence, \: it \: is \: solved \: !}$

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Answer 2

Answer:

n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not divisible by 3. n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3. thus one and only one out of n , n+2, n+4 is divisible by 3

Step-by-step explanation:

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