The Concept of the number system is very vast, and so is its application which examiners take advantage in setting up the papers. One has to prepare these concepts in such a way that it will develop the confidence to solve problems. Now we are going to discuss about the concept of a unit digit in below.
What is Unit Digit?
Numbers like 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are known as digits. When any number is expressed exponentially or by prime factorization we get a large number (Integer), so the digit at its unit or one’s position is known as the unit digit of the number.
Example:
Let N be any Number that can be expressed in
(a) N =2 x 1000 + 3 x 100 + 4x 10 + 6 x 1 = 2346, where 6 is the unit digit.
(b) N = 25 = 32, where 2 is the unit digit of the number.
(c) N = 25x 32 x 71 = 2016, where 6 is the unit digit.
- OPERATION BASED
- EXPONENTIAL BASED
- FACTORIAL BASED
OPERATION BASED (Unit digits after using operations)
It is always easy to locate the unit place digit of any operational number ( Addition and multiplication ) based.
Problem 01:
Find the unit digit of 267 + 564 + 768
Solution:
NOTE: Unit digit always depends on the digits at the unit place.
So, the entire number is not required to be added. Only focus on unit place digit and use the same operation as in the question ( 267 + 564 + 768 ).
Required Unit digit = 7 + 4 + 8 = 19, hence the unit digit is 9.
Problem 02:
Find the unit digit of 267 x 564 x 768
Solution:
Required answer = 7 x 4 x 8 = ( 7 x 4 ) x 8 = (28 ) x 8 = 8 x 8 = 64.
Hence, 4 is the unit digit.
NOTE: Always focus on the unit place digit only.
Problem 03:
The digit in the unit place of the product 81 x 82 x 83 x …..x 98 x 99 is
Solution:
As per the question the number is 81 x 82 x 83 x …..x 98 x 99.
We can write the number as 81 x 82 x 83 x ……x 90 x…….x 98 x 99.
When we multiply any number by 0 then the resultant number always carries 0 at the unit place.
Hence, the unit digit of the number is 0.
EXPONENTIAL BASED
Finding the unit place digit of the exponential number (ab) is also easy if only if the process is followed. The process is widely known as the process of Cyclicity. Every unit digit has its own repetitive pattern when expressed in exponential form. The following table will help you to understand the pattern of cyclicity.
| Exponential Form of Number | Number | Unit Digit |
| 2 1 | 2 | 2 |
| 2 2 | 4 | 4 |
| 2 3 | 8 | 8 |
| 2 4 | 16 | 6 |
| 2 5 | 32 | 2 |
| 2 6 | 64 | 4 |
| 2 7 | 128 | 8 |
| 2 8 | 256 | 6 |
| 2 9 | 512 | 2 |
| 2 10 | 1024 | 4 |
| 2 11 | 2048 | 8 |
| 2 12 | 4096 | 6 |
One can observe after every 4th power, the cycle of the unit digit repeats ( i.e 2, 4, 8, 6).
The following table shows the cycles of the natural numbers from 1 to 9:
| Natural Number (digit at the base) | Cycle | Unit digit based on the Cyclic Pattern |
| 1 | 1 | 1 |
| 2 | 4 | 2, 4, 8, and 6 |
| 3 | 4 | 3, 9, 7, and 1 |
| 4 | 2 | 4 and 6 |
| 5 | 1 | 5 |
| 6 | 1 | 6 |
| 7 | 4 | 7, 9, 3, and 1 |
| 8 | 4 | 8, 4, 2, and 6 |
| 9 | 2 | 9 and 1 |
From the table, it is observed that there are 3 such numbers whose unit digit will be the same, come what the power raised to the number. The numbers are ( 1, 5, and 6 ).
Problem 04:
Find the Unit Digit of 231343
Solution:
The unit digit for the number will always be 1
Problem 05:
Find the Unit Digit of 2534^54
Solution:
As in the expression, 5 is the unit digit of the base thus 5 is the answer.
From the table, it is observed that there are 2 such numbers whose cyclicity is two. ( i.e 4 and 9)
- 4odd: When the power raised to 4 is odd, the unit digit will always be 4.
- 4even: When the power raised to 4 is even, the unit digit will always be 6.
- 9odd: When the power raised to 9 is odd, the unit digit will always be 9.
- 9even: When the power raised to 9 is even, the unit digit will always be 1.
Problem 06:
What will be the unit digit of (4444)1024?
Solution:
The unit digit of the number (4444)1024is equal to the unit digit of (4)1024. The power raised to 4 is even. Thus the unit digit is 6.
From the table, it is observed that there are 4 such numbers whose cyclicity is four. ( i.e 2, 3, 7, and 8)
Problem 07:
The last digit in the expansion of 2256 is
Solution:
Working steps
Step 1: Divide the power or index by the cyclicity of the base number ( i.e 4).
Step 2: Note the remainder
Step 3: The remainder raised to the base gives the required unit digit.
The cyclicity of 2 is 4. So, dividing 256 by 4 we get 64 complete cycles and 0 as the remainder.
Now, 24 gives 6 as a unit digit.
so, the required unit digit of the number is 6
Problem 08:
Find the Unit Digit of 1734
Solution:
The unit digit of the number (17)34 is equal to the unit digit of (7)34.
Working steps
Step 1: Divide the power or index by the cyclicity of the base number (i.e 4)
Step 2: Note the remainder.
Step 3: The remainder raised to the base gives the required unit digit.
The cyclicity of 7 is 4. So, dividing 34 by 4 we get 8 complete cycles and 2 as the remainder.
72 gives 9 as a unit digit.
So, the required unit digit of the number is 9
Problem 09:
Find the Unit Digit of 3333 x 4444
Solution:
Note: The unit digit depends upon the unit place number. So, the Unit digit of 3333 is as same as 333.
Working steps:
Step 1: Divide the power or index by the cyclicity of the base number (i.e 4).
Step 2: Note the remainder.
Step 3: The remainder raised to the base gives the required unit digit.
For base 3: The cyclicity of 3 is 4. So, dividing 33 by 4 we get 8 complete cycles and 1 as the remainder.
31 gives 3 as a unit digit. So, the required unit digit of the number is 3.
For base 4: Power is even, the unit digit will be 6.
The required unit digit of 3333 x 4444 is 3 x 6 = 18, i.e 8 is the unit digit.
FACTORIAL BASED
The factorial of number N can be represented as ( N ! ) is the form of a number where the product of the natural numbers up to N.
N ! = N x ( N -1) x ( N – 2) x ( N – 3) x ……..x 2 x 1
Problem 10:
Find the last digit of 99 !
Solution:
we can write 99! = 99 x 98 x 97 x ……x 3 x 2 x 1
There are many numbers of the above expression whose unit digit is zero (0). When zero (0) is multiplied by any number it will give zero as a unit digit.
Problem 11:
The last digit of the expression 1! + 2! + 3! + 4! + ….. + 10! Is
Solution:
The above problem is just for testing the observation one’s concept of factorial.
We know, 1! = 1
2! = 2 x 1 = 2
3! = 3 x 2 x 1 = 6
4! = 4 x 3 x 2 x1 = 24
5! = 5 x 4 x 3 x 2 x1 = 120
Henceforth the unit digit of the numbers greater than or equal to 5 is zero ( 0 ).
So, Required unit digit of 1! + 2! + 3! + 4! + ….. + 10!
= 1 + 2 + 6 + 4 + 0
= 13
The unit digit will be 3
Problem 12:
A prime number contains the digit X at the unit’s place. How many such digits of X are possible?
Solution:
We know 2 is the only even prime number and no other prime number has an even number at the unit digit.
Thus no. of digits occupy at the unit digit of prime number = 6 ( i.e 1, 2, 3, 5, 7, and 9).
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