Unit Digit MCQ Quiz - Objective Question with Answer for Unit Digit - Download Free PDF

Given  

Given expression :  (234)100 + (234)101 

Concept Used 

For any natural number of having unit digit 4, 

If the power is even number i.e. 2, 4, 6, ...

Then the unit digit will be 6

And if the power is odd number i.e. 1, 3, 5, ...

Then the unit digit will be 4

Calculation 

Given expression is (234)100 + (234)101 

⇒ (234)¹⁰⁰[1 + 234]

⇒ (234)100 × 235

Here the power is 100(even) and number is 234(unit digit is 4)

⇒ The unit digit will be 6

Now, 6 × 5 = 30 (Unit digit is 0) 

∴ The correct answer is 0.

Given:

The product of {(341)491 × (625)317 × (6374)1793} and we are to find its unit digit.

Formula used:

To find the unit digit of a product, we multiply the unit digits of the individual numbers raised to their respective powers.

Calculation:

For (341)⁴⁹¹, the unit digit of 341 is 1. Any number ending in 1 raised to any power will have a unit digit of 1.

For (625)³¹⁷, the unit digit of 625 is 5. Any number ending in 5 raised to any power will have a unit digit of 5.

For (6374)¹⁷⁹³, the unit digit of 6374 is 4. The unit digit of numbers ending in 4 alternates in a cycle of 2 when raised to powers: 4¹ has a unit digit of 4, and 4² has a unit digit of 6, then it repeats.

Since 1793 is odd, 4 raised to an odd power will have a unit digit of 4.

Now, 1 (from 341⁴⁹¹) × 5 (from 625³¹⁷) × 4 (from 6374¹⁷⁹³) = 20

∴ The unit digit in the given product is 0.

Given:

We need to find the unit digit of (593)²³ × (124)²⁶.

Calculation:

First, look at the unit digit of (593)²³:

593 ends in 3. The cycle of powers for 3 is 3, 9, 7, 1, repeating every 4 powers.
23 mod 4

⇒ 3 (since 23 divided by 4 leaves a remainder of 3).

So, the unit digit of (593)²³ is 7 (the third number in the cycle).

Next, look at the unit digit of (124)²⁶:

124 ends in 4. The unit digit of the powers of 4 alternates between 4 and 6 every second power.

26 is even, so ⇒ the unit digit of (124)²⁶ is 6.

Therefore, the unit digit of the product (593)23 × (124)26  is found by multiplying the unit digits of each component (7 × 6):

Since the unit digit of 7 × 6 is 2, the unit digit of (593)²³ × (124)²⁶ is 2.

Given:

What is the digit in the units place of the number (1829)⁴²?

Formula used:

To find the unit digit of a number raised to a power, we use the cyclicity of the unit digits in powers of the base number.

Calculation:

The unit digit of 1829 is 9.

Unit digits of powers of 9 follow a cyclic pattern:

9¹ = 9 → Unit digit = 9

9² = 81 → Unit digit = 1

9³ = 729 → Unit digit = 9

9⁴ = 6561 → Unit digit = 1

Cyclic pattern: 9, 1

Now, divide the exponent (42) by the length of the cycle (2) to find the remainder:

⇒ 42 ÷ 2 = 21 remainder 0

When the remainder is 0, the unit digit corresponds to the last number in the cycle.

From the cycle (9, 1), the last number is 1.

∴ The digit in the units place of (1829)⁴² is 1.

Given:

Number = (1624)²⁴

Formula Used:

The digit in the units place of a power of a number depends only on the digit in the units place of the base.

The cycle of the units digits for powers of 4 is:

4¹ = 4

4² = 16 (units digit is 6)

4³ = 64 (units digit is 4)

4⁴ = 256 (units digit is 6)

The pattern of the units digit of powers of 4 is 4, 6, 4, 6, ...

If the exponent is odd, the units digit is 4.

If the exponent is even, the units digit is 6.

Calculation:

The units digit of the base (1624) is 4.

The exponent is 24, which is an even number.

According to the pattern of the units digits for powers of 4, when the exponent is even, the units digit of the result is 6.

Therefore, the units digit of (1624)²⁴ is 6.

∴ The digit in the units place of the number (1624)²⁴ is 6.

Given:

(432)⁴¹² × (499)⁴³¹

Concept:

9^{even no.} = unit digit 1

9^{odd no.} = unit digit 9

Calculation:

(432)⁴¹² × (499)⁴³¹

Taking unit digits

⇒ 2⁴¹² × 9⁴³¹

As we know unit digit of 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 6

⇒ 2⁴⁽¹⁰³⁾ × 9⁴³¹

⇒ 6 × 9

⇒ 54

∴ The unit digit of (432)⁴¹² × (499)⁴³¹ is 4.

To determine the last digit of the number 432412, we need to focus on the last digit of base 432 i.e. 2 and the exponential part 412.

We know,

Notice the pattern of the last digit. It is 2, 4, 8, 6, 2, 4, 8, 6, 2 …… so on.

Thus the last digit is repetitive and is a four-digit long i.e. 1, 2, 8, 6. If we keep on writing this table till the power of 2 reaches 412 then how many times this pattern repeated can be found by dividing 412 by 4.

412 divided by 4 is 103 with remainder 0 which indicates that the pattern gets fully repeated 412 times and then ends up with the digit i.e. 4. (if it is fully divisible we take power as 4)

∴ The Last digit of the number 432412 is 6.

9even no. = unit digit 1

9odd no. = unit digit 9

∴ The Last digit of the number 9431 is 9

∴ The unit digit of (432)412 × (499)431 is 4.

Unit digit of (164)¹⁶⁹ + (333)³³⁷ – (727)⁷²⁶

To check unit place divide power by 4

4¹⁶⁹ + 3³³⁷ – 7⁷²⁶

⇒ 69/4 = Reminder 1

⇒ 37/4 = Reminder 1

⇒ 26/4 = Reminder 2

⇒ 4¹ + 3¹ – 7²

⇒ 4 + 3 – 9

⇒ 7 – 9

or, 17 – 9

⇒ 8

Computing the factorials,

⇒ 1! = 1

⇒ 2! = 1 × 2 = 2

⇒ 3! = 1 × 2 × 3 = 6

⇒ 4! = 1 × 2 × 3 × 4 = 24

⇒ 5! = 1 × 2 × 3 × 4 × 5 = 120

As we can notice that the unit digit of 5! Is 0, hence, the unit digit of 6!, 7!, …, 50! will be zero

Unit's digit of (1! + 2! + 3! + 4! + …………… + 50!)

⇒ Unit's digit of (1! + 2! + 3! + 4!)

⇒ Unit's digit of (1 + 2 + 6 + 24)

∴ Unit's digit of (33) = 3

Alternate Method 1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

∴ Unit digit = 1+ 2+ 6 + 4 + 0  = 13 = 3  

Given:

(795 – 358) 

Concept used:

Cyclicity of 7 is 4

Cyclicity of 3 is 4

Calculation:

7⁹⁵ = 7^{(4 × 23) + 3} = 1 × 7³ = 343

⇒ Unit digit of 7⁹⁵ = 3

3⁵⁸ = 3^{(4 × 14) + 2} = 1 × 3² = 9

⇒ Unit digit of 3⁵⁸ = 9

Unit digit of (7⁹⁵ – 3⁵⁸) = 3 – 9 = -6 or 10 - 6 = 4

∴ Unit place will be 4

Note: 3 - 9 = -6 because we have to find the unit digit. whenever the result is negative add 10 in it.

Concept Used:

Factorial is defined as:

n! = n × (n - 1) × (n - 2) × … × 3 × 2 × 1

Calculation:

1! = 1 = 1 ⇒ Unit digit = 1

2! = 2 × 1 = 2 ⇒ Unit digit = 2

3! = 3 × 2 × 1 = 6 ⇒ Unit digit = 6

4! = 4 × 3 × 2 × 1 = 24 ⇒ Unit digit = 4

5! = 5 × 4 × 3 × 2 × 1 = 120 ⇒ Unit digit = 0

6! = 6 × 5! = 720 ⇒ Unit digit = 0

Hence, the required unit digit is 3.

Calculation:

The unit place of 7¹ = 7, 7² = 9, 7³ = 3, 7⁴ = 1

The unit place of 795  = 7^{23 × 4} × 7³ = 3

The unit place of 3¹ = 3, 3² = 9, 3³ = 7, 3⁴ = 1

The unit place of 358 = 3^{14 × 4} × 3² = 9

The unit of 795 is 3, which is less than 9

Then take 3 has 13 (by carry rule)

The unit place of N = 795 - 358 = 13 - 9 = 4

∴ The unit digit of N is 4

The unit digit of 25⁴³ is 5.

The unit digit of 56⁴² is 6.

The unit digit of 456²⁵ is 6.

The unit digit of 23⁴² is 9.

The unit digit of 76²³ is 6.

∴ The resultant value of the unit digits = [(5 × 6) + 6 + 9 + 6] = (30 + 6 + 9 + 6) = 51

So, the unit digit of the expression is 1.

Important Points

23⁴²: Here, we need to know that the unit place of powers of 3 repeats after every 4th power.

So, we divide the power with 4 and check the value of the remainder.

42/4 → 2 (remainder)

So, the unit digit will be 3² = 9.

The unit digits of other numbers end with 5 and 6, the unit digit of the power of which are the number itself throughout.

Concept:

To find the unit digit, consider the unit digit of the given number and find its cyclicity from the power.

Example: 2⁵

∵ 2 as a unit digit repeats after every 4 power, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32

⇒ Cyclicity of 2 = 4

Unit digit of 2⁵ = 2¹ = 2

Similarly, cyclicity of 3 = 4

Calculation:

Unit digit of (1373)³⁶ – (1442)²⁰

⇒ Unit digit of (3)³⁶ – (2)²⁰

⇒ Unit digit of {(3)^{9 × 4} – (2)^{5 × 4}}

⇒ Unit digit of {(3)⁴ – (2)⁴}

⇒ Unit digit of {81 – 16}

⇒ Unit digit of 65

Concept:

The unit digit of power of a number repeats itself every 4th time

Calculation:

353 - 638 + 2756

⇒ (3)(13 × 4) + 1 - (6)(9 × 4) + 2 + (27)(14 × 4)

⇒ Unit digit will be determined by unit digit of (31 - 6² + 274) 

⇒ Unit digit of 31 = 3

⇒ Unit digit of 62 = 6

⇒ Unit digit of 274 = 1

Hence, unit digit will be (3 - 6 + 1) = - 2

Since unit digit can not be negative, we will add 10 to it because these numbers are in decimal form.

∴ Unit digit of (3)53 - (6)38 + (27)⁵⁶ = - 2 + 10 = 8

Given:

3674 × 8596 + 5699 × 1589

Concept used:

The digit placed at the unit place is also called the digit at one's place.

Calculation:

To find unit digit multiply their unit digits = 4 × 6 + 9 × 9 

To find unit digit add their unit digits = 24 + 81 

So, unit digit is = 4 + 1 = 5