What is the unit digit of the expression K = (233^{85} \times 437^{94}) - 168^{47}? — State PSC Exam

What is the unit digit of the expression $K = (233^{85} \times 437^{94}) - 168^{47}$ ?

Answer: 5

The unit digit of the given expression is 5.

To find the unit digit of

K = (233^{85} \times 437^{94}) - 168^{47}

, analyze each term using base cyclicity rules:
1.

233^{85}

: Base digit 3 has cyclicity 4.

85 \div 4 = 21

remainder 1.

3^1 = 3

.
2.

437^{94}

: Base digit 7 has cyclicity 4.

94 \div 4 = 23

remainder 2.

7^2 = 49 \rightarrow 9

.
3. Product unit digit:

3 \times 9 = 27 \rightarrow 7

.
4.

168^{47}

: Base digit 8 has cyclicity 4.

47 \div 4 = 11

remainder 3.

8^3 = 512 \rightarrow 2

.
5. Subtracting gives

7 - 2 = 5

Find the unit digit of

233^{85}

Unit digit is 3

The unit digit of powers of 3 follows a cycle of 4: (3, 9, 7, 1). Since

85 \equiv 1 \pmod 4

, the unit digit is

3^1 = 3

Find the unit digit of

437^{94}

Unit digit is 9

The unit digit of powers of 7 follows a cycle of 4: (7, 9, 3, 1). Since

94 \equiv 2 \pmod 4

, the unit digit is

7^2 = 49 \implies 9

Find the unit digit of the product

(233^{85} \times 437^{94})

Unit digit is 7

Multiplying the unit digits of the two terms gives

3 \times 9 = 27

, which has a unit digit of 7.

Find the unit digit of

168^{47}

Unit digit is 2

The unit digit of powers of 8 follows a cycle of 4: (8, 4, 2, 6). Since

47 \equiv 3 \pmod 4

, the unit digit is

8^3 = 512 \implies 2

Calculate the final unit digit of the composite expression

Subtracting the unit digit of the subtracted term from the product's unit digit yields

7 - 2 = 5

Unit Digit and Cyclicity

Estimated Time:1m 30s

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