Question

Show that one and only one out of n, n+2 or n +4 is divisible by 3, where n is a
positive integer.​

Answer 1

Answer:

n can be taken as any positive integer... So n = 1

n+2 = 1+2 = 3... 3 is divisible by 3...

n+4 = 1+4 = 5... 5 is not divisible by 3...

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Answer 2

GIVEN:

• n, n+2, n+4

TO FIND:

• Prove that one and only one out of n, n+2 or n +4 is divisible by 3

SOLUTION:

We have given that, n is any positive integer.

According to Euclid's Division Lemma:-

CASE:- 1

Let n = 3

Put the value of 3 in positive integers

$\rm{\dashrightarrow n = 3 \: (Divisible \: by \: 3) }$

$\rm{\dashrightarrow n + 2 = 3+2 = 5 }$

$\rm{\dashrightarrow n +4 = 3+4 = 7}$

Here, we can see that only n is divisible by 3

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CASE:- 2

In integer n + 2, Put 3 at the place of n

➩ 3 + 2 = 5

$\rm{\dashrightarrow n = 3+2 = 5 }$

$\rm{\dashrightarrow n+2 = 3+ 2 + 2 = 5+2 = 7 }$

$\rm{\dashrightarrow n+4 = 3+2 + 4 = 5+4 = 9 \: (Divisible \: by \: 3) }$

In this case, we can see that, only n+4 is divisible by 3

CASE:- 3

In integer n + 4, Put 3 at the place of n

➩ 3 + 4 = 7

$\rm{\dashrightarrow n = 3+4 = 7}$

$\rm{\dashrightarrow n +2 = 3+ 4 + 2 = 7+2 = 9 \: (Divisible \: by \: 3) }$

$\rm{\dashrightarrow n +4 = 3+4 + 4 = 7+4 = 11 }$

In this case, we can see that, only (n + 2) is divisible by 3

❝ Hence, in each case only one out of n, n+2 or n +4 is divisible by 3 ❞

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