What is the unit digit of the composite exponential expression E = (238^{440} + 547^{321})… — State PSC Exam

What is the unit digit of the composite exponential expression $E = (238^{440} + 547^{321}) \times 929^{103}$ ?

The unit digit of the given expression is 7.

The unit digit of

238^{440}

is 6 (since

440 \pmod 4 = 0 \implies 8^4 \rightarrow 6

). The unit digit of

547^{321}

is 7 (since

321 \pmod 4 = 1 \implies 7^1 \rightarrow 7

). Their sum has a unit digit of

(6 + 7) \rightarrow 3

. The unit digit of

929^{103}

is 9 (since 103 is odd

\implies 9^1 \rightarrow 9

). Multiplying the components gives

3 \times 9 = 27

, yielding a final unit digit of 7.

Find the unit digit of

238^{440}

The unit digit of base 238 is 8. The cyclicity of 8 is 4 (

8, 4, 2, 6

). Divide the exponent 440 by 4:

440 \pmod 4 = 0

. Since the remainder is 0, we take the 4th power in the cycle (

8^4

), which ends in 6.

When an exponent is completely divisible by the cycle length (remainder 0), the position corresponds to the maximum cycle power (power 4).

Find the unit digit of

547^{321}

The unit digit of base 547 is 7. The cyclicity of 7 is 4 (

7, 9, 3, 1

). Divide exponent 321 by 4:

321 \pmod 4 = 1

. The 1st power in the cycle (

7^1

) ends in 7.

The unit digit is determined by the remainder of the exponent modulo 4.

Calculate the unit digit of the expression inside the parentheses:

(238^{440} + 547^{321})

Unit digit =

(6 + 7) \pmod{10} = 13 \pmod{10} = 3

Addition of unit digits follows standard modular arithmetic.

Find the unit digit of

929^{103}

The unit digit of base 929 is 9. The cyclicity of 9 is 2 (

9, 1

). Since the exponent 103 is odd,

9^{\text{odd}}

ends in 9.

Bases ending in 9 have an alternating cyclicity of 2.

Compute the final unit digit of

E = (3 \times 9)

Unit digit =

(3 \times 9) \pmod{10} = 27 \pmod{10} = 7

Multiplying the resultant unit digits gives the final unit digit of the composite product.

Unit digit cyclicity rule and exponent modulo operations

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