NIMCET 2015 — Reasoning PYQ

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Question

NIMCET 2015 — Reasoning PYQ

NIMCET | Reasoning | 2015

The remainder when $2^{31}$ is divided by $5$ is:

Choose the correct answer:

Explanation

Method 1: Cyclicity of Unit Digit

The remainder when a number is divided by $5$ depends on its unit digit. Let's look at the powers of $2$ :

$2^{1} = 2$
$2^{2} = 4$
$2^{3} = 8$
$2^{4} = 16$ (Unit digit is $6$ )
$2^{5} = 32$ (Unit digit is $2$ again)

The unit digits follow a cycle of $4$ : $\{2, 4, 8, 6\}$ .

To find the unit digit of $2^{31}$ :

Divide the power by $4$ : $31 \div 4 = \text{Remainder } 3$ .
The unit digit of $2^{31}$ will be the same as $2^{3}$ , which is $8$ .
When $8$ is divided by $5$ , the remainder is:

$8 \div 5 \implies \text{Remainder } = 3$

Method 2: Using Modular Arithmetic

We can find a power of $2$ that is close to a multiple of $5$ :

We know that $2^{2} = 4$ , and $4 \equiv -1 \pmod 5$ .

Now, express $2^{31}$ in terms of $2^{2}$ :

2^{31} = 2^{30} \times 2^{1}

2^{31} = (2^{2})^{15} \times 2

Substituting the remainder:

2^{31} \equiv (-1)^{15} \times 2 \pmod 5

2^{31} \equiv -1 \times 2 \pmod 5

2^{31} \equiv -2 \pmod 5

Since a remainder cannot be negative, add the divisor ( $5$ ):

-2 + 5 = 3

Conclusion:

The remainder when $2^{31}$ is divided by $5$ is $3$ .

Correct Option: (c) 3