show that exactly one of the numbers n,n+2 or n+4 is divisible by 3
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Answered by
38
Solution: We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur
Case I : When n = 3q
In this case, we have,
n=3q, which is divisible by 3
n=3q
= adding 2 on both sides
n + 2 = 3q + 2
n + 2 leaves a remainder 2 when divided by 3
Therefore, n + 2 is not divisible by 3
n = 3q
n + 4 = 3q + 4 = 3(q + 1) + 1
n + 4 leaves a 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 a reaminder 1 when divided by 3
n is not divisible by 3
n = 3q + 1
n + 2 = (3q + 1) + 2 = 3(q + 1)
n + 2 is divisible by 3
n = 3q + 1
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2
n + 4 leaves a 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
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 divsible by 3
n = 3q + 2
n + 4 = 3q + 2 + 4 = 3(q + 2)
n + 4 is divisible by 3
Thus, n + 4 is divisble by 3 but n and n + 2 are not divisible by 3
Case I : When n = 3q
In this case, we have,
n=3q, which is divisible by 3
n=3q
= adding 2 on both sides
n + 2 = 3q + 2
n + 2 leaves a remainder 2 when divided by 3
Therefore, n + 2 is not divisible by 3
n = 3q
n + 4 = 3q + 4 = 3(q + 1) + 1
n + 4 leaves a 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 a reaminder 1 when divided by 3
n is not divisible by 3
n = 3q + 1
n + 2 = (3q + 1) + 2 = 3(q + 1)
n + 2 is divisible by 3
n = 3q + 1
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2
n + 4 leaves a 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
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 divsible by 3
n = 3q + 2
n + 4 = 3q + 2 + 4 = 3(q + 2)
n + 4 is divisible by 3
Thus, n + 4 is divisble by 3 but n and n + 2 are not divisible by 3
pavi23:
very good
Answered by
18
Solution:
let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
remainders may be 0,1 and 2
n may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n 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 AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE
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