When a positive integer n is divided by 9, the remainder is 4. What is the remainder when (3n3 + 2n2 + 5n) is divided by 9?

When a positive integer n is divided by 9, the remainder is 4. What is the remainder when (3n3 + 2n2 + 5n) is divided by 9? Correct Answer 1

Given: 

When a positive integer n is divided by 9, the remainder is 4.

Concept used:

Dividend = Divisor × Quotient + Remainder

Calculation:

According to the concept,

3n3 is divided by 9 will leave

⇒ 3 × 4 × 4 × 4

⇒ 192

⇒ when 192 is divided by 9 it leaves a remainder of 3.

2n2 is divided by 9 will leave

⇒ 2 × 4 × 4

⇒ 32

⇒ when 32 is divided by 9 it leaves a remainder of 5.

In the case of 5n,

when 5 is divided by 9 leaves a remainder of 5.

So, when 5n is divided by 9, it will leave a remainder of

⇒ 5 × 4

⇒ 20

⇒ when 20 is divided by 9 it leaves a remainder of 2.

Let the quotients be P, Q, and R respectively.

Now, when (3n3 + 2n2 + 5n) is divided by 9

⇒ (9P + 3) + (9Q + 5) + (9R + 2)

⇒ 9(P + Q + R) + 10

⇒ 9(P + Q + R + 1) + 1

⇒ Remainder of 1

∴ 1 is the remainder when (3n3 + 2n2 + 5n) is divided by 9.

Related Questions