Understanding the Remainder Problem

The question asks us to find the remainder when the square of a specific number is divided by 5. We are given a condition about the original number: when this number is divided by 5, the remainder is 3.

Mathematical Representation

Let the number be represented by '$N$'. According to the problem statement, when '$N$' is divided by 5, the remainder is 3. We can express this using modular arithmetic as:

$N \equiv 3 \pmod{5}$

This means that '$N$' can be written in the form '$N = 5k + 3$', where '$k$' is any integer (like 0, 1, 2, ...).

Calculating the Square of the Number

We need to find the remainder when '$N^2$' (the square of the number) is divided by 5. Let's find '$N^2$':

$N^2 = (5k + 3)^2$

Expanding this expression:

$N^2 = (5k)^2 + 2 \cdot (5k) \cdot 3 + 3^2$

$N^2 = 25k^2 + 30k + 9$

Finding the Remainder of the Square

Now, we need to find the remainder when '$N^2$' is divided by 5. Let's look at the expression '$25k^2 + 30k + 9$' and consider its remainder when divided by 5:

• The term '$25k^2$' is a multiple of 5 (since 25 is $5 \times 5$), so its remainder when divided by 5 is 0.
• The term '$30k$' is also a multiple of 5 (since 30 is $5 \times 6$), so its remainder when divided by 5 is 0.
• The term '9' leaves a remainder when divided by 5.

So, we can write:

$N^2 \pmod{5} \equiv (25k^2 + 30k + 9) \pmod{5}$

$N^2 \pmod{5} \equiv (0 + 0 + 9) \pmod{5}$

$N^2 \pmod{5} \equiv 9 \pmod{5}$

Final Remainder Calculation

To find the final remainder, we divide 9 by 5:

$9 = 5 \times 1 + 4$

The remainder is 4.

Therefore, the remainder when the square of the number is divided by 5 is 4.

Alternative Method using Modular Arithmetic

We know that $N \equiv 3 \pmod{5}$.

To find the remainder of $N^2$ when divided by 5, we can square both sides of the congruence:

$N^2 \equiv 3^2 \pmod{5}$

$N^2 \equiv 9 \pmod{5}$

Since $9$ divided by $5$ leaves a remainder of $4$ ($9 = 5 \times 1 + 4$), we have:

$N^2 \equiv 4 \pmod{5}$

This confirms that the remainder is 4.