Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
Answers
Case-I: When n = 3q
In 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..
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Step-by-step explanation:
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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