Question

show that only one of every three consecutive positive integers is divisible by 3

Answer 1

● Let n,n+1,n+2 be three consecutive positive integers.

● We know that n is of the form 3q,3q+1 or, 3q+2

So, we have the following

Case l When n= 3q:

In this case, n is divisible by 3 but n+1 and n+2 are not divisible by 3.

Case II When n=3q+1:

In this case, n+2=3q+1+2=3 is divisible by 3 but n and n+1 are not divisible by 3.

Case III When n=3q+2:

In this case, n+1=3q+1+2=3(q+1) is divisible by 3 but n and n+2 are not divisible by 3.

Hence one of n,n+1 and n+2 is divisible by 3.

Answer 2

Answer:

yes they are divisible

Step-by-step explanation:

assuming the 3 consecutive numbers as

n,n+1,n+2

when a number is divided by 3 we can get 0,1,2 as remainder(basic math)

so the number can take the  form of 3p,3p+1,and 3p+2

if we take n as 3p

then n is divisible by 3

if we  take

n=3p+1

then

n+2 =3p+1+2

=3p+3

=3(p+1)

is divisible by 3

if n=3p+2

then

n+1=3p+2+1

3(p+1)

is divisible by 3

hence anyone can say that one among the number n,n+1,n+2 are always divisible by 3

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