Question

Show that only one of numbers n,n+2,n+3 is divisible by 3

Answer 1

n this case, we have

n= 3q, which is divisible by 3

Now, n = 3q

n+2 = 3q+2

n+2 leaves remainder 2 when divided by 3

Again, n = 3q

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

n+4 leaves remainder 1 when divided by 3

n+4 is not divisible by 3.

Thus, n is divisible by 3 but n+2 and n+4 are not divisible by 3.

 

Case-II: when n = 3q+1

In this case, we have

n= 3q+1,

n leaves remainder 1 when divided by 3.

n is divisible by 3

Now, n = 3q+1

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

n+2 is divisible by 3.

Again, n = 3q+1

n+4 = 3q+1+4=3q+5=3(q+1)+2

n+4 leaves remainder 2 when divided by 3

n+4 is not divisible by 3.

Thus, n+2 is divisible by 3 but n and n+4 are not divisible by 3.

 

Case-III: When n + 3q+2

In this case, we have

n= 3q+2

n leaves remainder 2 when divided by 3.

n is not divisible by 3.

Now, n = 3q+2

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

n+2 leaves remainder 1 when divided by 3

n+2 is not divisible by 3.

Again, n = 3q+2

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

n+4 is divisible by 3.

Thus, n+4 is divisible by 3 but n and n+2 are not divisible by 3

Answer 2

Step-by-step explanation:

[ There is some error in your question , it will be n + 4 instead of n + 3 . ]

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

thus any number is in the form of 3q , 3q+1 or 3q+2.

case I: if n =3q

n = 3q = 3(q) is divisible by 3,

n + 2 = 3q + 2 is not divisible by 3.

n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

case II: if n =3q + 1

n = 3q + 1 is not divisible by 3.

n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.

n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.

case III: if n = 3q + 2

n =3q + 2 is not divisible by 3.

n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

thus one and only one out of n , n+2, n+4 is divisible by 3.

Hence, it is solved

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