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.