Question

Show that exactly 1 of nos.....

Answer 1

Let the number n be divided by 3 gives q as quotient and r as remainder .

By Euclid's division lemma

a = bq + r

0 ≤ r < b


n = 3q + r

So , r = 0 , 1 , 2

n = 3q

n = 3q + 1

n = 3q + 2


Now ,

☢ Case :- 1 ( a )

n = 3q

It is divisible by 3 leaving remainder 0.

☣ Case :- 1 ( b )

n = 3q

n + 2 = 3q + 2

It is not divisible by 3 because it leave 2 as remainder.

☣ Case :- 1 ( c )

n = 3q

n + 4 = 3q + 4

It is not divisible by 3 because it leaves 4 as remainder.

☢ Case :- 2 ( a )

n = 3q + 1

It is not divisible by 3 because it leaves 1 as remainder.

☣ Case :- 2 ( b )

n = 3q + 1

n + 2 = 3q + 1 + 3

n + 2 = 3q + 3

It is divisible by 3 because it leaves remainder 3 which can further divide to give 0.

☣ Case :- 2 ( c )

n = 3q + 1

n + 4 = 3q + 1 + 4

n + 4 = 3q + 5

It is not divisible by 3 because it leaves 5 as remainder.


☢ Case :- 3 ( a )

n = 3q + 2

It is not divisible by 3 because it leaves 2 as remainder.

☣ Case :- 3 ( b )

n = 3q + 2

n + 2 = 3q + 2 + 2

n + 2 = 3q + 4

It is not divisible by 3 because it leaves 4 as remainder.

☣ Case :- 3 ( c )

n = 3q + 2

n = 3q + 2 + 4

n = 3q + 6

It is divisible by 3 because it leaves 6 as remainder which further can divide by 3 to get 0 as remainder.

Hence proved only one of them are divisible by 3 .

Answer 2

HEYA ,
HERE IS YOUR ANSWER,
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Let n be any +ve integer, then
n = 3q+r, r = 0,1,2
n = 3p or 3q+1 or 3q+2

● CASE I :-
when, n = 3q, which is not divisible by 3
n+2 = 3q+2, which is not divisible by 3
n+4 = 3q+4, which is not divisible by 3

●CASE II :-
When, n = 3q+1, which is not divisible by 3
n+2 = 3q+1+2 = 3q+3, which is divisible by 3
n+4 = 3q+1+4 = 3q+5, which is not divisible by 3

●CASE III :-
When, n = 3q+2, which is not divisible by 3
n+2 = 3q+2+2 = 3q+4, which is not divisible by 3
n+4 = 3q+2+4 = 3q+6, which is divisible by 3

☆Case I , II and III ==> One and only one out of new, n+2 or n+4 is divisible by 3.
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#HOPE IT HELPS U !!!!