Question

Show that only one of the numbers n,n+2and n+4 is divisible by 3.

Answer 1

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.

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Answer 2

Let n be an integer and b = 3n = 3 q + r

Where q is qoutient and r is remainder

0....
$< r < 3$
So,

the remainder may be 0, 1 and 2

so n may be in the form of 3q , 3q + 1 and

3q + 2

CASE 1

if n = 3q

n + 4 = 3q + 4

n + 2 = 3q+ 2

here now is only divisible by 3


CASE 2

if n = 3q + 1

n + 4 = 3q + 5

n + 2 = 3q = 3

here only n + 2 is divisible by 3

CASE 3

if n = 3q + 2

n + 2= 3q + 4

n + 4 = 3q + 2 + 4

=3q + 6

here only n + 4 is divisible by 3

........Hence,

it is justified that one and only one amoung n, n+ 2 and n + 4. is divisible by 3 in each case.......