Unit Digit and Cyclicity Practice Questions — State PSC Exam

What is the unit digit of the expression $7^{43} + 9^{27}$ ?

Show answer & explanation

Answer: 2

Answer

The unit digit of the given expression

7^{43} + 9^{27}

is 2.

The unit digit of

7^{43}

is determined by dividing 43 by 4, giving a remainder of 3, so

7^3

ends in 3. The unit digit of

9^{27}

is determined by taking 9 to an odd power, which ends in 9. Adding

3 + 9 = 12

gives a final unit digit of 2.

Step-by-Step Solution

1

Determine the unit digit of

7^{43}

using cyclicity rules.

The unit digit of

7^{43}

is 3.

The base 7 follows a 4-step cyclicity pattern (7, 9, 3, 1). Dividing exponent 43 by 4 leaves remainder 3, giving

7^3 \pmod{10} = 3

.

2

Determine the unit digit of

9^{27}

using cyclicity rules.

The unit digit of

9^{27}

is 9.

The base 9 follows a 2-step cyclicity pattern (9 for odd powers, 1 for even powers). Since 27 is odd, the unit digit is 9.

3

Combine the resulting unit digits.

The unit digit of the sum is 2.

Summing the individual unit digits gives

3 + 9 = 12

, whose unit digit is 2.

Key Concept

Unit Digit and Cyclicity

Question 2Question →

What is the unit digit of the expression $S = (1! + 2! + 3! + \dots + 99!) + 3^{99}$ ?

Show answer & explanation

Answer: 0

Answer

The unit digit of the given expression is 0.

The unit digit of a sum is determined by the sum of the unit digits of its individual terms. For the factorial sum

(1! + 2! + 3! + \dots + 99!)

, terms from

5!

onward end in 0 because

5! = 120

. Summing the first four terms yields

1 + 2 + 6 + 24 = 33

, contributing a unit digit of 3. For

3^{99}

, the cyclicity sequence of base 3 is 3, 9, 7, 1 (length 4). Dividing 99 by 4 leaves a remainder of 3, corresponding to

3^3 = 27

, which contributes a unit digit of 7. Summing the two unit digits yields

3 + 7 = 10

, giving a final unit digit of 0.

Step-by-Step Solution

1

Evaluate the unit digit contribution of the factorial terms

Unit digit of

(1! + 2! + 3! + \dots + 99!)

is 3

Since

n!

ends in 0 for every

n \ge 5

, only

1! + 2! + 3! + 4! = 33

affects the unit digit.

2

Find the unit digit of

3^{99}

using pattern cyclicity

Unit digit of

3^{99}

is 7

The cyclicity of base 3 is 4. Since

99 \pmod 4 = 3

, the unit digit corresponds to

3^3 = 27

.

3

Combine the unit digits of both parts

Unit digit of

S

is 0

Adding the unit digits gives

3 + 7 = 10

, making the final unit digit 0.

Key Concept

Unit digit evaluation using factorial terminal zeros and exponential cyclicity

Question 3Question →

Determine the unit digit of the composite exponential expression $E = 17^{18^{19}} + 18^{19^{20}} - 19^{20^{21}}$ .

Show answer & explanation

Answer: 8

Answer

8

The unit digit of

17^{18^{19}}

is 1 because

18^{19}

is divisible by 4, which corresponds to the 4th power

7^4 \equiv 1 \pmod{10}

. The unit digit of

18^{19^{20}}

is 8 because

19^{20} \equiv 1 \pmod 4

, giving

8^1 = 8

. The unit digit of

19^{20^{21}}

is 1 because

20^{21}

is an even exponent, giving

9^{\text{even}} \equiv 1 \pmod{10}

. Combining these gives

1 + 8 - 1 = 8

.

Step-by-Step Solution

1

Find the unit digit of

17^{18^{19}}

The unit digit is 1

Base 7 has cyclicity 4. Exponent

18^{19} = 2^{19} \equiv 0 \pmod 4

. When the remainder modulo 4 is 0, we take the 4th power in the cycle:

7^4 \equiv 1 \pmod{10}

.

2

Find the unit digit of

18^{19^{20}}

The unit digit is 8

Base 8 has cyclicity 4. Exponent

19^{20} \equiv (-1)^{20} \equiv 1 \pmod 4

. The 1st power in the cycle gives

8^1 = 8

.

3

Find the unit digit of

19^{20^{21}}

The unit digit is 1

Base 9 has cyclicity 2. The exponent

20^{21}

is even, so

9^{\text{even}} \equiv 1 \pmod{10}

.

4

Evaluate the combined expression

The final unit digit is 8

Combine the unit digits of the individual terms:

1 + 8 - 1 = 8

.

Key Concept

Cyclicity of numbers and modular arithmetic for tower exponents

Question 4Question →

What is the unit digit of the expression $2^{52} + 5^{17}$ ?

1

6

7

3

Show answer & explanation

Answer: 1

Answer

The unit digit of the expression

2^{52} + 5^{17}

is 1.

For

2^{52}

, dividing 52 by 4 leaves a remainder of 0. Since the cyclicity of 2 is 4, a remainder of 0 means using

2^4 = 16

, which has a unit digit of 6. For

5^{17}

, any power of 5 ends in 5. The total unit digit is the unit digit of

6 + 5 = 11

, which is 1.

Step-by-Step Solution

1

Find the unit digit of

2^{52}

using cyclicity.

The cyclicity of 2 is 4 (pattern: 2, 4, 8, 6). Dividing the exponent 52 by 4 yields

52 = 4 \times 13 + 0

. A remainder of 0 corresponds to the 4th power in the cycle.

2^4 = 16

, so the unit digit of

2^{52}

is 6.

When the exponent is exactly divisible by the cyclicity length, the unit digit corresponds to the power equal to the cyclicity length.

2

Find the unit digit of

5^{17}

.

The cyclicity of 5 is 1, meaning any positive integer power of 5 ends in 5. Thus, the unit digit of

5^{17}

is 5.

Base numbers ending in 5 always yield a unit digit of 5 for any positive integer exponent.

3

Add the individual unit digits to get the final unit digit.

Unit digit of

(2^{52} + 5^{17}) = \text{Unit digit of }(6 + 5) = \text{Unit digit of }11 = 1

.

The unit digit of a sum is equal to the unit digit of the sum of the individual unit digits.

Key Concept

Unit Digit and Cyclicity

Estimated Time:45s

Question 5Question →

What is the unit digit of the expression $E = (842^{324} \times 617^{153}) + 358^{217}$ ?

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Answer: 0

Answer

The unit digit of the expression

E = (842^{324} \times 617^{153}) + 358^{217}

is 0.

For

842^{324}

, exponent

324 \pmod 4 = 0

corresponds to the 4th power position in cyclicity (

2^4 = 16 \rightarrow 6

). For

617^{153}

,

153 \pmod 4 = 1

corresponds to

7^1 = 7

. The product of these terms ends in

6 \times 7 = 42 \rightarrow 2

. For

358^{217}

,

217 \pmod 4 = 1

corresponds to

8^1 = 8

. Adding the product's unit digit (2) and

358^{217}

's unit digit (8) gives

2 + 8 = 10 \rightarrow 0

.

Step-by-Step Solution

1

Find the unit digit of

842^{324}

Unit digit is 6

The unit digit of base 842 is 2, which repeats unit digits in a cycle of 4:

(2, 4, 8, 6)

. Dividing exponent 324 by 4 yields remainder 0, indicating the 4th position in the cycle,

2^4 = 16 \rightarrow 6

.

2

Find the unit digit of

617^{153}

Unit digit is 7

The unit digit of base 617 is 7, which repeats in a cycle of 4:

(7, 9, 3, 1)

. Dividing exponent 153 by 4 yields remainder 1, indicating the 1st position in the cycle,

7^1 = 7

.

3

Calculate the unit digit of the product

(842^{324} \times 617^{153})

Unit digit is 2

Multiplying the unit digits of the two terms gives

6 \times 7 = 42

, which has a unit digit of 2.

4

Find the unit digit of

358^{217}

Unit digit is 8

The unit digit of base 358 is 8, which repeats in a cycle of 4:

(8, 4, 2, 6)

. Dividing exponent 217 by 4 yields remainder 1, indicating the 1st position in the cycle,

8^1 = 8

.

5

Add the unit digit of the product to the unit digit of

358^{217}

Final unit digit is 0

Adding the unit digit of the product (2) and the unit digit of the final term (8) gives

2 + 8 = 10

, which ends in 0.

Key Concept

Unit Digit Computation using Exponents and Cyclicity

Question 6Question →

In basic arithmetic, the unit digit of a power of an integer repeats in a periodic pattern known as cyclicity. What is the unit digit of $3^{45}$ ?

Show answer & explanation

Answer: 3

Answer

The unit digit of

3^{45}

is 3.

The base number 3 has a cyclicity pattern of length 4: (3, 9, 7, 1). When the exponent 45 is divided by 4, the quotient is 11 and the remainder is 1. A remainder of 1 indicates the unit digit is equal to the first element of the cyclic pattern, which is 3.

Step-by-Step Solution

1

Identify the unit digit cyclicity pattern for base 3.

The unit digits repeat every 4 powers in the pattern (3, 9, 7, 1).

Calculating sequential powers of 3 yields last digits of 3, 9, 7, 1, 3, 9, 7, 1, establishing a cycle length of 4.

2

Divide the exponent by the cycle period of 4.

45 \div 4 = 11

with a remainder of

1

.

The remainder indicates the exact position within the repeating cycle.

3

Find the unit digit using the remainder.

Since the remainder is 1, the unit digit is

3^1 = 3

.

The first number in the cyclicity sequence (3, 9, 7, 1) is 3.

Key Concept

Unit Digit Cyclicity

Estimated Time:45s

Question 7Question →

What is the unit digit of the composite exponential expression $E = (432^{316} \times 657^{235}) + 875^{432} - 959^{411}$ ?

4

9

6

0

Show answer & explanation

Answer: 4

Answer

The unit digit of the expression is 4.

Evaluating each term using cyclicity:

432^{316}

ends in 6 (

2^4

),

657^{235}

ends in 3 (

7^3

),

875^{432}

ends in 5 (

5^n

), and

959^{411}

ends in 9 (

9^{\text{odd}}

). Combining these gives

(6 \times 3) + 5 - 9 = 18 + 5 - 9 = 14

, yielding a unit digit of 4.

Step-by-Step Solution

1

Determine the unit digit of

432^{316}

Unit digit is 6

The unit digit of the base is 2, which has a cyclicity pattern of 4 (2, 4, 8, 6). The exponent 316 is divisible by 4 with remainder 0. For a remainder of 0, we take the 4th power in the cycle:

2^4 = 16

, giving a unit digit of 6.

2

Determine the unit digit of

657^{235}

Unit digit is 3

The unit digit of the base is 7, which has a cyclicity pattern of 4 (7, 9, 3, 1). Dividing the exponent 235 by 4 gives a remainder of 3 (

235 = 4 \times 58 + 3

). We take the 3rd power in the cycle:

7^3 = 343

, giving a unit digit of 3.

3

Calculate the unit digit of the product

(432^{316} \times 657^{235})

Unit digit is 8

Multiplying the unit digits of the two terms yields

6 \times 3 = 18

, which has a unit digit of 8.

4

Determine the unit digit of

875^{432}

and add it to the product

Unit digit of sum is 3

Any positive integer power of a base ending in 5 always ends in 5. Adding this to the product's unit digit gives

8 + 5 = 13

, so the sum ends in 3.

5

Determine the unit digit of

959^{411}

and subtract it to find the final unit digit

Final unit digit is 4

The unit digit of the base is 9, which has a cyclicity of 2 (9 for odd powers, 1 for even powers). Since 411 is odd,

959^{411}

ends in 9. Subtracting this from the sum's unit digit gives

3 - 9 \equiv 13 - 9 = 4

.

Key Concept

Unit digit calculation using cyclicity patterns and modular arithmetic for multi-term exponential expressions.

Question 8Question →

What is the unit digit of the expression $8^{20} + 7^{13}$ ?

1

3

5

8

Show answer & explanation

Answer: 3

Answer

The unit digit of the given expression is 3.

The unit digit of

8^{20}

is 6 because 20 is a multiple of 4, matching the 4th term in the cyclicity cycle (8, 4, 2, 6). The unit digit of

7^{13}

is 7 because 13 leaves a remainder of 1 when divided by 4, matching the 1st term in the cyclicity cycle (7, 9, 3, 1). Adding these unit digits gives

6 + 7 = 13

, whose unit digit is 3.

Step-by-Step Solution

1

Find the unit digit of

8^{20}

Unit digit is 6

The cyclicity of numbers ending in 8 is 4 (pattern: 8, 4, 2, 6). Since 20 is divisible by 4 (

20 \pmod 4 = 0

), the unit digit corresponds to the 4th power in the cycle, which is 6.

2

Find the unit digit of

7^{13}

Unit digit is 7

The cyclicity of numbers ending in 7 is 4 (pattern: 7, 9, 3, 1). Dividing 13 by 4 gives a remainder of 1 (

13 \pmod 4 = 1

), so the unit digit corresponds to

7^1 = 7

.

3

Add the unit digits of both terms

3

Sum of unit digits is

6 + 7 = 13

. Taking the unit digit of 13 gives 3.

Key Concept

Cyclicity of Unit Digits

Estimated Time:45s

Question 9Question →

What is the unit digit of the composite exponential expression $S = 717^{100} + 424^{101} - 818^{100}$ ?

9

4

1

7

Show answer & explanation

Answer: 9

Answer

The unit digit of the expression is 9.

Evaluating each term using its base cyclicity gives unit digits of 1 for

717^{100}

, 4 for

424^{101}

, and 6 for

818^{100}

. Combining them yields

1 + 4 - 6 = -1

. Adding 10 to adjust for borrowing in unit digit subtraction gives

-1 + 10 = 9

.

Step-by-Step Solution

1

Find the unit digit of

717^{100}

Unit digit is 1

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

7^1=7, 7^2=9, 7^3=3, 7^4=1

). Dividing the exponent 100 by 4 leaves remainder 0, which corresponds to the 4th power:

7^4 \equiv 1 \pmod{10}

.

2

Find the unit digit of

424^{101}

Unit digit is 4

The unit digit of the base is 4. The cyclicity of 4 is 2 (

4^{\text{odd}}=4, 4^{\text{even}}=6

). Since the exponent 101 is odd, the unit digit is 4.

3

Find the unit digit of

818^{100}

Unit digit is 6

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

8^1=8, 8^2=4, 8^3=2, 8^4=6

). Dividing the exponent 100 by 4 leaves remainder 0, which corresponds to the 4th power:

8^4 \equiv 6 \pmod{10}

.

4

Combine the unit digits to find the final result

Unit digit is 9

Substitute the individual unit digits into the expression:

1 + 4 - 6 = 5 - 6 = -1

. Converting

-1

to a positive unit digit modulo 10 gives

-1 + 10 = 9

.

Key Concept

Unit digit determination using cyclicity and modular arithmetic for composite exponential expressions.

Estimated Time:1m 15s

Question 10Question →

Find the unit digit of the numerical expression $N = (567^{63} \times 234^{45}) - 343^{37}$ .

Show answer & explanation

Answer: 9

Answer

9

The unit digit of

567^{63}

is 3 and for

234^{45}

it is 4, making their product's unit digit 2. The unit digit of

343^{37}

is 3. Subtracting 3 from 2 with regrouping (

12 - 3

) yields 9.

Step-by-Step Solution

1

Determine the unit digit of

567^{63}

3

Base unit digit is 7 with cyclicity 4. Since

63 \pmod 4 = 3

,

7^3

gives a unit digit of 3.

2

Determine the unit digit of

234^{45}

4

Base unit digit is 4 with cyclicity 2. An odd exponent yields a unit digit of 4.

3

Multiply the unit digits of the first two terms

2

The unit digit of the product is

(3 \times 4) \pmod{10} = 2

.

4

Determine the unit digit of

343^{37}

3

Base unit digit is 3 with cyclicity 4. Since

37 \pmod 4 = 1

,

3^1

gives a unit digit of 3.

5

Subtract the unit digit of the second part from the first part

9

Subtracting 3 from 2 requires borrowing 10 (

12 - 3 = 9

) to yield a valid positive unit digit.

Key Concept

Unit Digit and Cyclicity of Powers

Question 11Question →

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

7

2

9

1

Show answer & explanation

Answer: 7

Answer

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.

Step-by-Step Solution

1

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).

2

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.

3

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.

4

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.

5

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.

Key Concept

Unit digit cyclicity rule and exponent modulo operations

Estimated Time:2m 0s

Question 12Question →

What is the unit digit of the expression $432^{216} - 517^{142} - 734^{121} \times 283^{105}$ ?

5

0

1

4

Show answer & explanation

Answer: 5

Answer

The unit digit of the given expression is 5.

Evaluating each component:

432^{216}

ends in 6 (since 216 is divisible by 4, giving

2^4 = 16

),

517^{142}

ends in 9 (

142 \pmod 4 = 2

, giving

7^2 = 49

), and

734^{121} \times 283^{105}

ends in

4 \times 3 = 12 \rightarrow 2

. Combining these gives

(6 - 9 - 2) = -5 \equiv 5 \pmod{10}

. Thus, the option with value 5 is correct.

Step-by-Step Solution

1

Determine the unit digit of the first term

432^{216}

Unit digit is 6

The base unit digit is 2, which has a cyclicity of 4 (pattern: 2, 4, 8, 6). Dividing exponent 216 by 4 gives a remainder of 0. When the remainder is 0, we use the 4th power:

2^4 = 16

, so the unit digit is 6.

2

Determine the unit digit of the second term

517^{142}

Unit digit is 9

The base unit digit is 7, which has a cyclicity of 4 (pattern: 7, 9, 3, 1). Dividing exponent 142 by 4 gives a remainder of 2. Taking

7^2 = 49

, the unit digit is 9.

3

Determine the unit digit of the product term

734^{121} \times 283^{105}

Unit digit is 2

For

734^{121}

, base 4 with an odd exponent gives unit digit 4. For

283^{105}

, base 3 with exponent

105 \pmod 4 = 1

gives unit digit 3. Multiplying their unit digits:

4 \times 3 = 12

, so the unit digit is 2.

4

Combine the unit digits following order of operations (BODMAS)

Unit digit is 5

We compute

(6 - 9 - 2) \pmod{10}

. Evaluating left-to-right:

6 - 9 = -3 \equiv 7 \pmod{10}

(by adding 10 to handle borrowing). Then

7 - 2 = 5

.

Key Concept

Unit digit calculation using cyclicity of numbers and modular arithmetic rules under BODMAS

Estimated Time:2m 0s

Question 13Question →

What is the unit digit of the expression $E = 138^{124} + 264^{93} - 177^{54}$ ?

1

6

9

3

Show answer & explanation

Answer: 1

Answer

The unit digit of the expression is 1.

The correct answer is derived by determining the cyclicity of each base's unit digit: 8 has cyclicity 4 (with remainder 0 yielding unit digit 6), 4 has cyclicity 2 (odd exponent yielding unit digit 4), and 7 has cyclicity 4 (remainder 2 yielding unit digit 9). Combining these yields (6 + 4) - 9 = 1.

Step-by-Step Solution

1

Find the unit digit of

138^{124}

Unit digit is 6

The unit digit of the base is 8. The cyclicity of 8 is 4 (8, 4, 2, 6). Dividing the exponent 124 by 4 gives a remainder of 0. A remainder of 0 corresponds to the 4th power in the cycle (

8^4 = 4096

), so the unit digit is 6.

2

Find the unit digit of

264^{93}

Unit digit is 4

The unit digit of the base is 4. The cyclicity of 4 is 2 (

4^1 = 4, 4^2 = 6

). Since the exponent 93 is odd, the unit digit is 4.

3

Find the unit digit of

177^{54}

Unit digit is 9

The unit digit of the base is 7. The cyclicity of 7 is 4 (7, 9, 3, 1). Dividing the exponent 54 by 4 gives a remainder of 2 (

54 = 4 \times 13 + 2

). Thus, the unit digit is

7^2 = 49

, which ends in 9.

4

Combine the unit digits according to the expression

Unit digit is 1

Evaluating

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

. The unit digit of the entire expression is 1.

Key Concept

Unit Digit Cyclicity and Modular Exponentiation Rules

Question 14Question →

What is the unit digit of the composite exponential expression $E = 332^{520} \times 553^{213} - 448^{631}$ ?

1

0

6

4

Show answer & explanation

Answer: 6

Answer

The unit digit of the given expression is 6.

Evaluating each component using cyclicity of unit digits:

332^{520}

has base unit digit 2 and exponent divisible by 4 (

520 \bmod 4 = 0

), so its unit digit is

2^4 \implies 6

.

553^{213}

has base unit digit 3 with

213 \bmod 4 = 1

, giving unit digit

3^1 = 3

.

448^{631}

has base unit digit 8 with

631 \bmod 4 = 3

, giving unit digit

8^3 \implies 2

. Combining these gives

(6 \times 3) - 2 = 18 - 2 = 16

, which yields a final unit digit of 6.

Step-by-Step Solution

1

Determine the unit digit of

332^{520}

using cyclicity of 2

Unit digit is 6

The unit digit of base 332 is 2. The cyclicity of 2 is 4 (2, 4, 8, 6). Dividing the exponent 520 by 4 gives a remainder of 0. When remainder is 0, we take the 4th power:

2^4 = 16

, so the unit digit is 6.

2

Determine the unit digit of

553^{213}

using cyclicity of 3

Unit digit is 3

The unit digit of base 553 is 3. The cyclicity of 3 is 4 (3, 9, 7, 1). Dividing the exponent 213 by 4 gives a remainder of 1 (

213 = 4 \times 53 + 1

). Thus, the unit digit is

3^1 = 3

.

3

Determine the unit digit of

448^{631}

using cyclicity of 8

Unit digit is 2

The unit digit of base 448 is 8. The cyclicity of 8 is 4 (8, 4, 2, 6). Dividing the exponent 631 by 4 gives a remainder of 3 (

631 = 4 \times 157 + 3

). Thus, the unit digit is

8^3 = 512

, which ends in 2.

4

Combine the unit digits according to the expression

E = (332^{520} \times 553^{213}) - 448^{631}

Unit digit is 6

First multiply the unit digits of the first two terms:

6 \times 3 = 18 \implies 8

. Then subtract the unit digit of the third term:

8 - 2 = 6

.

Key Concept

Unit Digit Cyclicity Rules

Estimated Time:1m 30s

Question 15Question →

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

Show answer & explanation

Answer: 5

Answer

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

.

Step-by-Step Solution

1

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

.

2

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

.

3

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.

4

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

.

5

Calculate the final unit digit of the composite expression

5

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

7 - 2 = 5

.

Key Concept

Unit Digit and Cyclicity

Estimated Time:1m 30s

Question 16Question →

Determine the unit digit of the finite series and exponent tower expression $S = \sum_{k=1}^{50} (k!)^{k!} + 7^{7^7}$ .

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Answer: 0

Answer

The unit digit of the expression is 0.

The sum of the factorial terms from k=1 to k=4 contributes unit digits of 1, 4, 6, and 6, totaling 17 (unit digit 7). Terms for k >= 5 end in 0. The power tower term 7^(7^7) has an exponent 7^7 which leaves a remainder of 3 when divided by 4. Since the cyclicity of 7 is 4, 7^3 gives a unit digit of 3. Adding 7 and 3 results in 10, giving a final unit digit of 0.

Step-by-Step Solution

1

Calculate unit digits of factorial terms for k from 1 to 4

Unit digits are 1, 4, 6, and 6 respectively

1! = 1, 2! = 2, 3! = 6, 4! = 24 (with base ending in 4 raised to an even power 24 giving unit digit 6)

2

Analyze factorial terms for k >= 5

Unit digit is 0 for all k >= 5

Factorials for k >= 5 contain factors 2 and 5, making the trailing digit 0

3

Sum the unit digits of the series sum

Sum of unit digits is 1 + 4 + 6 + 6 = 17, giving unit digit 7

Only terms from k = 1 to 4 contribute to the unit digit of the factorial sum

4

Find the unit digit of the power tower 7^(7^7) using cyclicity modulo 4

7^7 mod 4 = 3, so unit digit is 7^3 mod 10 = 3

The base 7 has a cyclicity of 4, and 7^7 is congruent to 3 modulo 4

5

Add the unit digits of both components

7 + 3 = 10, unit digit is 0

Combining the unit digit of the factorial sum (7) and the tower exponent term (3)

Key Concept

Combining factorial unit digit termination properties with exponent tower cyclicity modulo 4.

Question 17Question →

What is the unit digit of the composite expression $E = (1! + 2! + 3! + \dots + 20!)^{2026} + (3^1 \times 3^2 \times 3^3 \times \dots \times 3^{20})$ ?

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Answer: 8

Answer

The unit digit of the expression is 8.

Evaluating the expression requires breaking it down into two components. First, for the factorial sum

1! + 2! + 3! + \dots + 20!

, every term from

5!

onward contains factors of both 2 and 5, so its unit digit is 0. The unit digit of the sum is determined solely by

1! + 2! + 3! + 4! = 33

, which has a unit digit of 3. Raising 3 to the power 2026 gives

3^{2026}

. Since the unit digits of powers of 3 repeat in cycles of 4 (3, 9, 7, 1) and

2026 \equiv 2 \pmod 4

, the unit digit of

3^{2026}

is

3^2 = 9

.

Second, the product

3^1 \times 3^2 \times \dots \times 3^{20}

simplifies using the exponent addition rule to

3^{1+2+\dots+20} = 3^{210}

. Dividing 210 by 4 leaves a remainder of 2, so

3^{210}

also has a unit digit of

3^2 = 9

.

Adding the unit digits of both terms gives

9 + 9 = 18

, resulting in a final unit digit of 8.

Step-by-Step Solution

1

Find the unit digit of the inner factorial sum

S = 1! + 2! + 3! + \dots + 20!

.

The unit digit of

S

is 3.

For all

k \ge 5

,

k!

is divisible by 10 and ends in 0. Thus, only the sum of the first four terms

1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

determines the unit digit.

2

Calculate the unit digit of the first term

S^{2026} \equiv 3^{2026}

.

The unit digit of the first term is 9.

The unit digits of powers of 3 repeat in a cycle of 4 (3, 9, 7, 1). Dividing the exponent 2026 by 4 gives a remainder of 2 (

2026 = 4 \times 506 + 2

). Therefore, the unit digit is

3^2 = 9

.

3

Simplify the exponential product

P = 3^1 \times 3^2 \times 3^3 \times \dots \times 3^{20}

.

The product simplifies to

3^{210}

.

By exponent multiplication rules,

3^1 \times 3^2 \times \dots \times 3^{20} = 3^{\sum_{i=1}^{20} i}

. The sum of the first 20 positive integers is

\frac{20 \times 21}{2} = 210

.

4

Calculate the unit digit of

3^{210}

.

The unit digit of the second term is 9.

Dividing the exponent 210 by 4 gives a remainder of 2 (

210 = 4 \times 52 + 2

). Therefore, the unit digit is

3^2 = 9

.

5

Combine the unit digits of the two terms.

The unit digit of

E

is 8.

Adding the unit digits gives

9 + 9 = 18

. The unit digit of 18 is 8.

Key Concept

Unit Digit and Cyclicity of Factorial and Exponential Expressions

Estimated Time:2m 0s

Question 18Question →

What is the unit digit of the composite exponential expression $E = 433^{45} + 818^{80} - 625^{75}$ ?

4

9

2

6

Show answer & explanation

Answer: 4

Answer

4

Evaluating each term individually:

433^{45}

has unit digit

3

(since

45 \pmod 4 = 1

),

818^{80}

has unit digit

6

(since

80 \pmod 4 = 0

, giving the 4th power unit digit

8^4 \rightarrow 6

), and

625^{75}

has unit digit

5

. Combining these gives

3 + 6 - 5 = 4

.

Step-by-Step Solution

1

Find the unit digit of

433^{45}

Unit digit is 3

The unit digit of base 433 is 3. The cyclicity of 3 is 4 (

3^1=3, 3^2=9, 3^3=7, 3^4=1

). Dividing the exponent 45 by 4 gives

45 = 4 \times 11 + 1

(remainder 1). Thus, the unit digit is

3^1 = 3

.

2

Find the unit digit of

818^{80}

Unit digit is 6

The unit digit of base 818 is 8. The cyclicity of 8 is 4 (

8^1=8, 8^2=4, 8^3=2, 8^4=6

). Dividing exponent 80 by 4 gives a remainder of 0. When remainder is 0, we take the 4th power, giving unit digit

6

(from

8^4 = 4096

).

3

Find the unit digit of

625^{75}

Unit digit is 5

The unit digit of base 625 is 5. Any positive integer power of a number ending in 5 always ends in 5 (cyclicity of 1).

4

Combine the unit digits of all terms

4

Substitute the individual unit digits into the expression:

3 + 6 - 5 = 4

.

Key Concept

Unit Digit and Cyclicity Rules for Exponential Expressions

Question 19Question →

Determine the unit digit of the numerical expression $N = (333^{444} \times 444^{333}) + (777^{888} \times 888^{777}) - 222^{555}$ .

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Answer: 4

Answer

The unit digit of the given expression is 4.

By analyzing cyclicity of each base (

3, 4, 7, 8, 2

), we find that

333^{444} \rightarrow 1

,

444^{333} \rightarrow 4

,

777^{888} \rightarrow 1

,

888^{777} \rightarrow 8

, and

222^{555} \rightarrow 8

. Thus, the overall unit digit is

(1 \times 4) + (1 \times 8) - 8 = 4 + 8 - 8 = 4

.

Step-by-Step Solution

1

Calculate unit digit of

333^{444} \times 444^{333}

Unit digit is 4

Base 3 has cyclicity 4;

444 \pmod 4 = 0 \implies 3^4 \rightarrow 1

. Base 4 has cyclicity 2; odd exponent

333 \implies 4^1 \rightarrow 4

. Total unit digit for term 1 =

1 \times 4 = 4

.

2

Calculate unit digit of

777^{888} \times 888^{777}

Unit digit is 8

Base 7 has cyclicity 4;

888 \pmod 4 = 0 \implies 7^4 \rightarrow 1

. Base 8 has cyclicity 4;

777 \pmod 4 = 1 \implies 8^1 \rightarrow 8

. Total unit digit for term 2 =

1 \times 8 = 8

.

3

Calculate unit digit of

222^{555}

Unit digit is 8

Base 2 has cyclicity 4;

555 \pmod 4 = 3 \implies 2^3 \rightarrow 8

.

4

Evaluate combined expression unit digit

4

Combine term unit digits:

(4 + 8 - 8) = 4

.

Key Concept

Unit Digit and Cyclicity of Numbers

Question 20Question →

What is the unit digit of the numerical expression $N = (137^{137} \times 264^{102}) + 619^{199}$ ?

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Answer: 1

Answer

1

The unit digit of

137^{137}

is derived from

137 \pmod 4 = 1

, giving

7^1 = 7

. The unit digit of

264^{102}

is 6 because the exponent 102 is even. Multiplying these unit digits gives

7 \times 6 = 42

, which contributes a unit digit of 2. The unit digit of

619^{199}

is 9 because 199 is odd. Adding the unit digits yields

2 + 9 = 11

, whose unit digit is 1.

Step-by-Step Solution

1

Determine the unit digit of

137^{137}

using the cyclicity of 7.

The exponent 137 leaves a remainder of 1 when divided by 4, giving a unit digit of

7^1 = 7

.

The unit digit pattern for powers of 7 repeats every 4 powers (7, 9, 3, 1).

2

Determine the unit digit of

264^{102}

using the cyclicity of 4.

Since 102 is an even exponent,

4^{\text{even}}

yields a unit digit of 6.

Powers of 4 end in 4 for odd exponents and 6 for even exponents.

3

Calculate the unit digit of the product

(137^{137} \times 264^{102})

.

(7 \times 6) = 42

, which has a unit digit of 2.

The unit digit of a product depends only on the product of the unit digits of its factors.

4

Determine the unit digit of

619^{199}

using the cyclicity of 9.

Since 199 is an odd exponent,

9^{\text{odd}}

yields a unit digit of 9.

Powers of 9 end in 9 for odd exponents and 1 for even exponents.

5

Sum the unit digits to find the final unit digit of expression

N

.

2 + 9 = 11

, which has a unit digit of 1.

The unit digit of the sum of two terms is the unit digit of the sum of their individual unit digits.

Key Concept

Unit digit determination using exponent cyclicity rules for base digits 4, 7, and 9.

Estimated Time:1m 15s

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