Math, asked by sniper3638, 1 year ago

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

Answers

Answered by S4MAEL
32
hii Mate____☺

here's your answer_____⤵

☆We applied Euclid Division algorithm on n and 3.
a = bq +r  on putting a = n and b = 3
n = 3q +r 

● 0<r<3

●i.e n = 3q   -------- (1)

n = 3q +1 --------- (2)

n = 3q +2  -----------(3)

n = 3q is divisible by 3

●or n +2  = 3q +1+2 = 3q +3 also divisible by 3

●or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

_)======
☆HOPE IT HELPS!! ✔✔✔✔✌✌

sniper3638: Tq
S4MAEL: welcm dude
Answered by Anonymous
42
Hey Sniper,

Here is your solution :

Let , when n is divided by 3 then quotient is q and remainder is r.

Using Euclid's Division Lemma,

⇒ n = 3q + r         [ 0 ≤ r < 3 ]

So, possible values of r = 0,1,2.

In case 1,

when , r = 0.

⇒ n = 3q + 0

⇒ n = 3q ( Divisible )

By adding 2 to both sides,

⇒ n + 2 = 3q + 2 ( Not divisible )

By again adding 2 to both sides,

⇒ n + 2 + 2 = 3q + 2 + 2

⇒ n + 4 = 3q + 4

⇒ n + 4 = 3q + 3 + 1

⇒ n + 4 = 3 ( q + 1 ) + 1  ( Not divisible )

So, when r = o , then only one of the following is divisible.

Case 2.

When , r = 1.

Now,

⇒ n = 3q + 1 ( Not divisible )

By adding 2 to both sides ,

⇒ n + 2 = 3q + 1 + 2

⇒ n + 2 = 3q + 3

⇒ n + 2 = 3 ( q + 1 )    ( Divisible )

Again by adding 2 to both sides ,

⇒ n + 2 + 2 = 3q + 3 + 2

⇒ n + 4 = 3 ( q + 1 ) + 2             ( Not divisible )

In this case also only one of them is divisible .

Case 3,

When , r = 2.

Now,

⇒ n = 3q + 2 ( Not divisible )

By adding 2 to both sides ,

⇒ n + 2 = 3q + 2 + 2

⇒ n + 2 = 3q + 4

⇒ n + 2 = 3q + 3 + 1

⇒ n + 2 = 3 ( q + 1 ) + 1  ( Not divisible )

Again , by adding 2 to both sides ,

⇒ n + 2 + 2 = 3 ( q + 1 ) + 1 + 2

⇒ n + 4 = 3 ( q + 1 ) + 3

⇒ n + 4 = 3 ( q + 1 + 1 )

⇒ n + 4 = 3 ( q + 2 )    ( Divisible )

In this case also only one of them is divisible.

So, in all cases only one of those three is divisible by 3.


Hope it helps !!




Anonymous: Thanks Sniper for Brainliest
sniper3638: wlcm
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