Question

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

Answer 1

Answer:

make me as a brainlist

Step-by-step explanation:

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. hope this helps you.

Answer 2

Step-by-step explanation:

To solve this Question we must know the Euclids division Lemma or Division formula, both are the same thing but given different names that is

Dividend = Divisor × Quotient + Remainder

(This is always true let's check it through an example)

16 ÷ 3

_5__

3 | 16

-15

-----

1

16 = 3 × 5 + 1

But in division Lemma instead of writing the whole thing it is represented with alphabets

a = bq + r where 0 ≤ r < b (Think about it....) that is the remainders will be from 0 to number < b

where q is quotient and r is remainder, b is Divisor and a is dividend

1st case

a = n and b = 3 so r = 0, 1, 2

a = bq + r where 0 ≤ r < b

n = 3q + 0 = 3q

n = 3q + 1

n = 3q + 2

Here one and only 3q is where n is divisible by 3

2nd case

a = n + 2 and b = 3 so r = 0, 1, 2

a = bq + r where 0 ≤ r < b

n + 2 = 3q + 0 + 2 = 3q + 2

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

n + 2 = 3q + 2 + 2 = 3q + 4

Here one and only 3(q + 1) is divisible by 3

3rd case

a = n + 4 and b = 3 so r = 0, 1, 2

a = bq + r where 0 ≤ r < b

n + 4 = 3q + 0 + 4 = 3q + 4

n + 4 = 3q + 1 + 4 = 3q + 5

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2)

Here one and only 3(q + 2) is divisible by 3

Hence, if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3

We can also check this

let's take n = 5

n = 5 is not divisible by 3

n + 2 = 7 is not divisible by 3

n + 4 = 9 is divisible by 3

so,if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3

you can check this with any number

Hope you understood it........All the best