This question was previously asked in

UPSSSC Chakbandi Lekhpal Official Paper 4 (Held on : 1 Oct 2019 Shift 2)

Option 3 : either 0 or 1

Indian National Movement

101541

10 Questions
10 Marks
12 Mins

**Concept used:**

All the even numbers are in the form of 2q and all the odd numbers are in the form of 2q + 1 where q be any integer.

All the numbers divisible by 4 are in the form of 4k and all the numbers not divisible by 4 are in the form of 4k + m,

where k, m are integers and 0 < m < 4

If any number is in the form 4k + m, then m is the remainder when 4k + m is divided by 4.

**Calculation:**

Let any number be n.

The numbers are generally two types, either even or odd.

**Case - 1:**

Let n is even

According to the concept, n = 2q

Now, n^{2} = 4q^{2}

⇒ n2 = 4k

Here, n^{2} is multiple of 4.

As n^{2} is multiple 4, the remainder is 0 when n^{2} is divided by 4.

Case - 2:

Let n is odd

According to the concept, n = 2q + 1

Now, n2 = (2q + 1)2

⇒ 4q^{2} + 4q + 1

⇒ 4(q^{2} + q) + 1

⇒ 4k + 1

where k be any integer

Here, n2 is in the form of 4k + m, where m = 1.

So, the remainder is 1 when n2 is divided by 4.

**∴ The remainder is either 0 or 1 when the square of any number is divided by 4.**