What is the unit digit of the expression N = 2^{40} + 4^{31} - 7^{38}? — State PSC Exam

What is the unit digit of the expression $N = 2^{40} + 4^{31} - 7^{38}$ ?

The unit digit of the given expression is 1.

The unit digits of the individual terms are 6, 4, and 9 respectively. Adding the first two unit digits gives

6 + 4 = 10

(unit digit 0). Subtracting 9 from 10 yields

10 - 9 = 1

. Therefore, the overall unit digit of the expression is 1.

Find the unit digit of

2^{40}

using cyclicity.

The cyclicity of 2 is 4 (2, 4, 8, 6). Divide the exponent 40 by 4:

40 \pmod 4 = 0

. When the remainder is 0, take the 4th power:

2^4 = 16

, so the unit digit is 6.

Powers of 2 repeat their unit digits in cycles of 4.

Find the unit digit of

4^{31}

The cyclicity of 4 is 2 (4 for odd exponents, 6 for even exponents). Since 31 is odd, the unit digit is 4.

Odd powers of 4 always end in 4.

Find the unit digit of

7^{38}

The cyclicity of 7 is 4 (7, 9, 3, 1). Divide the exponent 38 by 4:

38 \pmod 4 = 2

7^2 = 49

, so the unit digit is 9.

A remainder of 2 corresponds to the second term in the cyclicity sequence of 7.

Combine the unit digits according to the expression

N = 2^{40} + 4^{31} - 7^{38}

Unit digit =

(6 + 4) - 9 = 10 - 9 = 1

Perform modular arithmetic modulo 10 to find the final unit digit.

Unit Digit Cyclicity and Modular Addition/Subtraction

Question