Question

show that one and only one out of n, n+2 or n+4 is divisible by 3 where n is any positive intiger

Answer 1

$\huge{\underline{\bf{\green{Solution:-}}}}$

On dividing n by 3 ,

Let q be the quotient and r be the remainder.

Then,

n = 3q + r , where 0 ≤ r < 3

➝ n = 3q + r , where r = 0,1,2

➝ n = 3q

➝ n = 3q +1

➝ n = 3q +2

Case I :-

If n = 3q then n is divisible by 3.

Case II :-

If n = 3q +1

then,

➝ ( n +2 ) = 3q +3

➝ 3(q+1),

which is divisible by 3.

So, (n +2 ) is divisible by 3.

Case III :-

When n = 3q +2

then (n+4) = 3q +6

➝ 3(q+2),

which is divisible by 3.

so, (n+4) is divisible by 3.

Hence,

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

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Answer 2

$\sf\green{\underline{\underline{To \ prove:}}}$

$\sf{One \ and \ only \ one \ out \ of \ n, \ n+2}$

$\sf{or \ n+4 \ is \ divisible \ by \ 3 \ where \ n}$

$\sf{is \ any \ positive \ integer.}$

$\bold\blue{\tt{\underline{\underline{Proof:}}}}$

$\sf\orange{Explanation:}$

$\sf{Multiples \ of \ 3 \ are}$

$\sf{3,6,9,12,...}$

$\sf{Here, \ it \ forms \ an \ A.P. \ with \ Common \ difference=3}$

$\sf{Any \ number \ divided \ by \ 3 \ gives}$

$\sf{0,1 \ or \ 2 \ as \ remainder \ only.}$

$\sf{\implies{If \ n \ is \ divided \ by \ 3 \ and \ we}}$

$\sf{get \ remainder \ 1. \ Then \ the \ number}$

$\sf{is \ to \ be \ perfectly \ divided \ if \ dividend\ was }$

$\sf{2 \ more, \ i.e. \ n+2.}$

$\sf{\implies{If \ n \ is \ divided \ by \ 3 \ perfectly}}$

$\sf{then \ it \ will \ be \ not \ divisible \ by \ n+2}$

$\sf{or \ n+4, \ because \ the \ common \ difference \ in }$

$\sf{numbers \ divided \ by \ 3 \ is \ of \ 3.}$

$\sf{\implies{If \ n \ is \ divided \ by \ 3 \ and \ we \ get}}$

$\sf{2 \ as \ remainder. \</p><p> \ The \ dividend \ will \ be \ perfectly}$

$\sf{divided \ if \ dividend \ was \ 1 \ more, \ i.e \ n+1}$

$\sf{But \ n+4 \ is \ such \ that \ (n+1)+3 \ thus \ it}$

$\sf{forms \ the \ next \ multiple \ of \ 3.}$

$\sf{Hence, \ n+4 \ will \ be \ perfectly \ divided.}$

_______________________________________

$\sf{e.g.}$

$\sf{(1) \ Let \ n \ be \ 6}$

$\sf{\therefore{n=6 \ and \ it's \ perfectly \ divided \ by \ 3.}}$

$\sf{\therefore{n+2=8, \ but \ i6ts \ not \ divided \ perfectly.}}$

$\sf{\therefore{n+4=10, \ but \ it's \ too \ not \ perfectly}}$

$\sf{divided \ by \ 3.}$

$\sf{(2) \ Let \ n \ be \ 4}$

$\sf{\therefore{n=4, \ but \ it's \ not \ perfectly \ divided \ by \ 3.}}$

$\sf{\therefore{n+2=6, \ it's \ perfectly \ divided \ by \ 3.}}$

$\sf{\therefore{n+4=10, \ but \ it's \ too \ not \ perfectly}}$

$\sf{divided \ by \ 3.}$

$\sf{(3) \ Let \ n \ be \ 2}$

$\sf{\therefore{n=2, \ it's \ not \ perfectly \ divided \ by \ 3.}}$

$\sf{\therefore{n+2=4, \ it's \ too \ not \ perfectly \ divided}}$

$\sf{by \ 3.}$

$\sf{\therefore{n+4=6, \ it's \ perfectly \ divided \ by \ 3.}}$

$\sf\purple{Hence, \ proved}$

$\sf\purple{\tt{One \ and \ only \ one \ out \ of \ n, \ n+2}}$

$\sf\purple{\tt{or \ n+4 \ is \ divisible \ by \ 3 \ where \ n}}$

$\sf\purple{\tt{is \ any \ positive \ integer.}}$