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