Q. 134.0( 12 Votes )

Show that exactly

Answer :

As per Euclid’s lemma,

If a and b are two positive integers, then they can be represented in the form,


a = bq + r


where 0 ≤ r < b


If b = 3, this can be written as


a = 3q + r


where 0 ≤ r < 3


So, r = 0, 1, 2


a = 3q, 3q + 1, 3q + 2


Let n = 3q, 3q + 1, 3q + 2


First taking n = 3q …(i)


This clearly implies n is divisible by 3.


Adding 2 on both sides of equation (i),


n + 2 = 3q + 2


Putting q = 1, we get


n + 2 = 3(1) + 2


n + 2 = 5


Now since 5 is not divisible by 3, this implies (n + 2) is not divisible by 3.


Adding 4 on both sides of equation (ii),


n + 4 = 3q + 4


Putting q = 1, we get


n + 4 = 3(1) + 4 = 7


Since 7 is not divisible by 3, this implies (n + 4) is not divisible by 3.


Now taking n = 3q + 1 …(ii)


Putting q =1, we get


n = 3(1) + 1 = 4


Since 4 is not divisible by 3, n is also not divisible by 3.


Adding 2 on both sides of equation (ii),


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


It’s clear that 3(q + 3), which is a multiple of 3, is divisible by 3.


(n + 2) is divisible by 3.


Adding 4 on both sides of equation (ii),


n + 4 = 3q + 1 + 4
= 3q + 5


Putting q = 1, we get


n + 4 = 3(1) + 5
= 3 + 5
= 8


8 is not divisible by 3.


Since (3q + 5) is not divisible by 3, this implies (n + 4) is also not divisible by 3.


Taking n = 3q + 2 …(iii)


Putting q = 1, we get


n = 3(1) + 2
= 5


5 is not divisible by 3.


(3q + 2) is not divisible by 3


n is not divisible by 3.


Adding 2 on both sides of equation (iii),


n + 2 = 3q + 2 + 2
= 3q + 4


Putting q = 1, we get


n + 2 = 3(1) + 4
= 7


Since 7 is not divisible by 3, this means (3q + 4) is not divisible by 3.


(n + 2) is not divisible by 3.


Adding 4 on both sides of equation (iii),


n + 4 = 3q + 2 + 4
= 3q + 6
= 3(q + 2)


Since 3(q + 2) is a multiple of 3, this means it is divisible by 3.


(n + 4) is divisible by 3.


Thus, we have seen exactly one of the numbers n, n + 2, n + 4 is divisible by 3.

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