What is the unit digit of the value obtained from the numerical expression N = (512^{160} … — State PSC Exam

What is the unit digit of the value obtained from the numerical expression $N = (512^{160} + 737^{203}) \times 434^{116} - 618^{140}$ ?

A
3
B
2
8Answer
D
4

Answer

Evaluating each term using cyclicity rules gives

512^{160} \rightarrow 6

737^{203} \rightarrow 3

434^{116} \rightarrow 6

, and

618^{140} \rightarrow 6

. The composite expression evaluates to

(6 + 3) \times 6 - 6 = 54 - 6

. Performing the unit digit subtraction

4 - 6

with borrowing yields

14 - 6 = 8

Step-by-Step Solution

Determine the unit digits of the terms inside the parentheses:

512^{160}

and

737^{203}

Unit digit of

512^{160}

6

; unit digit of

737^{203}

3

For

512

, the base unit digit is

2

(cyclicity pattern 2, 4, 8, 6). Exponent

160 \div 4

leaves remainder

0

, indicating the

4^{\text{th}}

power in cyclicity:

2^4 = 16 \rightarrow 6

. For

737

, the base unit digit is

7

(cyclicity pattern 7, 9, 3, 1). Exponent

203 \div 4

leaves remainder

3

, indicating

7^3 = 343 \rightarrow 3

Add the unit digits inside the parentheses and multiply by the unit digit of

434^{116}

Unit digit of the product term is

4

Sum inside parentheses:

6 + 3 = 9

. For

434^{116}

, the base unit digit is

4

(cyclicity pattern 4, 6). An even exponent

116

gives

4^2 = 16 \rightarrow 6

. The product is

9 \times 6 = 54 \rightarrow 4

Subtract the unit digit of

618^{140}

from the product's unit digit

The final unit digit is

8

For

618^{140}

, base unit digit is

8

(cyclicity pattern 8, 4, 2, 6). Exponent

140 \div 4

leaves remainder

0

, giving

8^4 = 4096 \rightarrow 6

. Subtracting gives

4 - 6

. Since

4 < 6

, add

10

to borrow from the tens place:

14 - 6 = 8

Key Concept

Cyclicity of Unit Digits and Modular Subtraction

Estimated Time:1m 30s

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