Question

N is a square of odd positive integer find the remainder when 5N + 3 is divided by 20

Answer 1

Answer:

The remainder is 8

Step-by-step explanation:

Given:

N is a square of odd positive integer

Then, $N=(2m+1)^2$ where m is a whole number

Now,

$5N+3$

$=5((2m+1)^2)+3$

$=5(4m^2+1+4m)+3$

$=20m^2+5+20m+3$

$=20m^2+20m+8$

$\implies\:N=20(m^2+m)+8$

Hence, when 5N+3 is divided by 20, the remainder is 8

Answer 2

Answer:

8

Step-by-step explanation:

Given that N is the square of any odd positive integer. We denote an odd positive integer as 2n + 1

So, N = (2n + 1)²

⇒ N = 4n² + 4n + 1

[using (a + b)² = a² + b² + 2ab]

We have to find the remainder when 5N + 3 is divided by 20.

Substitute the value of N in 5N + 3

20

5N+3

⟹

20

5(4n

2

+4n+1)+3

⟹

20

20n

2

+20n+5+3

⟹

20

20n

2

+20n+8

⟹

20

20(n

2

+n)+8

⟹

20

20(n

2

+n)

+

20

8

Clearly, 20n² + 20n would always be a multiple of 20. So the remainder would be given by the remaining term, that is 8. When 8 is divided by 20, the quotient is 0 and remainder is 8. Hence the answer is 8.

May it help you....