We know that, in various exams or while during calculations we come across the numbers with huge power and we have to find the last digit of that answer. It is very difficult to find but it takes time. In this article we learn how to find the last digit of power within seconds.
To learn it we have to find the cyclicity of numbers from 0 to 9 first.
Cyclicity of 0:
If there are numbers ending with zero and having huge powers then the last digit of that power will be always 0.
For example:
(m0)x= 0, where m is any digit number and x is any digit power.
(120)123=> last digit = 0
(100230)345 => last digit = 0
Similarly, (8904550)2356=> last digit= 0
Because any power of zero makes it zero always.
Hence, cyclicity of zero is one.
Cyclicity of 1:
We know that, if we have taken any power of 1 then the answer will be always one.
Hence, cyclicity of 1 is also one.
And any digit number ending with one with any digit power then last digit of power is one always.
For example:
(m1)x=> last digit = 1
Where, m is any digit number and x is the any digit power.
(12341)123=> last digit = 1
(8902341)3=> last digit= 1
(89991)234=> last digit= 1
Thus, any digit number ending with 1 and having any power then the last digit of its answer will be always 1.
Cyclicity of 5:
We know that, if we multiply 5 with 5 many times then the answer we got has the last digit always 5.
Hence, the cyclicity of 5 is also one.
The any digit number ending with 5 and having any digit power then the last digit of its answer will be always 5.
For example:
(m5)x => last digit=5
Where, m is the any digit number and x is the any digit power.
(25)23=> last digit= 5
(12345)234 => last digit = 5
(789045)3254=> last digit= 5
Thus, any digit number ending with 5 and having any digit power then the last digit of its answer is always 5.
Cyclicity of 6:
We know that, if we multiply 6 with 6 many times then we got the answer ending with 6 always.
Hence, cyclicity of 6 is also one.
Thus, any digit number ending with 6 and having any digit power then the last digit of its answer is always 6.
For example:
(m6)x => last digit= 6
Where m is any digit number and x is any digit power.
(6)2=> last digit=6
(36)23 => last digit=6
(890123456)789 => last digit= 6
Thus, any digit number ending with 6 and having any digit power then the last digit of its answer is always 6.
: Note:
As if we take any power of 0, 1, 5 and 6 then the last digit of its answer is always that number only.
Like (0)23 => last digit=0
(1)89 => last digit=1
(5)2 => last digit=5
(6)2 => last digit= 6
Thus, we can say that cyclicity of 0,1,5 and 6 is always one.
Cyclicity of 4:
We know that,
4*4= 42= 16 => last digit=6
4*4*4= 43 =64 => last digit=4
4*4*4*4= 44= 256 => last digit= 6
Similarly, (14)2 =196 => last digit=6
(14)3= 2744 => last digit=4
Thus, when the power of number ending with 4 is even then it’s last digit is 6.
When the power of number ending with 4 is odd then it’s last digit is 4.
That means, (m4)odd power=> last digit= 4
(m4)even power=> last digit=6
Thus, the cyclicity of 4 is two.
And power cycle becomes {4,6}.
Cyclicity of 9:
We know that,
9*9=92=81 => last digit=1
9*9*9= 93=729 => last digit= 9
9*9*9*9= 94 = 6561 => last digit=1
Similarly, (19)2= 361 => last digit=1
(19)3= 6859 => last digit=9
Thus, when the power of number ending with 9 is even then it’s last digit is 1.
And when power of number ending with 9 is odd then it’s last digit is 9.
That means, (m9)even power => last digit= 1
(m9)odd power => last digit= 9
Thus, the cyclicity of 9 is also two.
And it’s power cycle becomes {1,9}.
Note:
Cyclicity of 4 and 9 is two as it contains 2 elements.
Power cycle of 4 is {4, 6}.
Power cycle of 9 is {9, 1}.
Cyclicity of 2:
We know that,
21= 2 => last digit=2
2*2=22= 4 => last digit=4
2*2*2=23=8 => last digit= 8
24= 16 => last digit=6
25= 32 => last digit= 2
26=64 => last digit=4
27= 128 => last digit=8
28=256 => last digit=6
29= 512 => last digit= 2
In this way, if we take any power of 2 then it has last digit {2, 4, 8, 6} which is the cycle repeating and hence, power cycle of 2 is {2, 4, 8, 6}.
As it contains four numbers hence it’s cyclicity is 4.
For example:
1) What is the last digit of (122)46.
Here the number is ending with 2.
We know that, cyclicity of 2 is four and power cycle is {2, 4, 8, 6}.
Here power is 46.
So we divide power 46 by cyclicity 4, we get remainder as 2.
And 2nd number in power cycle is 4.
So last digit of (122)46 is 4.
If you check this answer on calculator then you will get the same answer.
2) Find the last digit of (2022)2020.
Here the number is ending with 2.
We know that, the cyclicity of 2 is four and it’s power cycle is {2,4,8,6}.
And here power is 2020 hence we divide power 2020 by cyclicity 4 then the we get the remainder as zero.
When remainder is zero then fourth element from the power cycle is the last digit.
In the power cycle of 2 the fourth element is 6.
Hence, here the last digit of (2022)2020 is 6.
Cyclicity of 3:
We know that,
31= 3 => last digit=3
3*3=32= 9 => last digit=9
3*3*3=33=27 => last digit= 7
34= 81 => last digit=1
35=243 => last digit=3
36=729 => last digit=9
37= 2187 => last digit=7
38=6561 => last digit=1
39= 19683 => last digit= 3
In this way, if we take any power of 3 then it has last digit {3, 9, 7, 1} which is the cycle repeating and hence, power cycle of 3 is {3, 9, 7, 1}.
As it contains four numbers hence it’s cyclicity is 4.
For example:
1) What is the last digit of (123)26
Here the number is ending with 3.
We know that, cyclicity of 3 is four and power cycle is {3, 9, 7, 1}.
Here power is 26.
So we divide power 26 by cyclicity 4, we get remainder as 2.
And 2nd number in power cycle of 3 is 9.
So last digit of (123)26 is 9.
If you check this answer on calculator then you will get the same answer.
2) Find the last digit of (2023)2021
Here the number is ending with 3.
We know that, the cyclicity of 3 is four and it’s power cycle is {3, 9, 7, 1}.
And here power is 2021. hence we divide power 2021 by cyclicity 4 then the we get the remainder as 1.
As the remainder is one the first element from the power cycle of 3 is 3.
Hence, here the last digit of (2023)2021 is 3.
Cyclicity of 7:
We know that,
71= 7 => last digit=7
7*7= 72= 49 => last digit=9
7*7*7=73=343 => last digit= 3
74=2401 => last digit=1
75=16807 => last digit=7
76=117649 => last digit=9
77= 823543 => last digit=3
78=5,764,801 => last digit=1
79=40,353,607 => last digit= 7
In this way, if we take any power of 7 then it has last digit {7, 9, 3, 1} which is the cycle repeating and hence, power cycle of 7 is {7, 9, 3, 1}.
As it contains four numbers hence it’s cyclicity is 4.
For example:
1) What is the last digit of (127)19
Here the number is ending with 7.
We know that, cyclicity of 7 is four and power cycle is {7, 9, 3, 1}.
Here power is 19.
So we divide power 19 by cyclicity 4, we get remainder as 3.
And 3rd number in power cycle of 7 is 3.
So last digit of (127)19 is 3.
If you check this answer on calculator then you will get the same answer.
2) Find the last digit of (2027)2022
Here the number is ending with 7.
We know that, the cyclicity of 7 is four and it’s power cycle is {7, 9, 3, 1}.
And here power is 2022 hence we divide power 2022 by cyclicity 4 then the we get the remainder as two.
As the remainder is 2 and hence second element from power cycle of 7 is 9.
Hence, here the last digit of (2027)2022 is 9.
Cyclicity of 8:
We know that,
81= 8 => last digit=8
8*8=82= 64 => last digit=4
8*8*8=83=512 => last digit= 2
84=4096 => last digit=6
85=32768 => last digit=8
86=262,144 => last digit=4
87= 2,097,152 => last digit=2
88=16,777,216 => last digit=6
89= 134,217,728 => last digit= 8
In this way, if we take any power of 8 then it has last digit {8, 4, 2, 6} which is the cycle repeating and hence, power cycle of 8 is {8, 4, 2, 6}.
As it contains four numbers hence it’s cyclicity is 4.
For example:
1) What is the last digit of (128)23
Here the number is ending with 8.
We know that, cyclicity of 8 is four and power cycle is {8, 4, 2, 6}.
Here power is 23.
So we divide power 23 by cyclicity 4, we get remainder as 3.
And 3rd number in power cycle of 8 is 2.
So last digit of (128)23 is 2.
If you check this answer on calculator then you will get the same answer.
2) Find the last digit of (2028)2020.
Here the number is ending with 8.
We know that, the cyclicity of 8 is four and it’s power cycle is {8, 4, 2, 6}.
And here power is 2020 hence we divide power 2020 by cyclicity 4 then the we get the remainder as zero.
When remainder is zero then fourth element from the power cycle of 8 is the last digit.
In the power cycle of 8 the fourth element is 6.
Hence, here the last digit of (2028)2020 is 6.
Note:
Cyclicity of 2, 3, 7 and 8 is four as their power cycle contains four elements.
Power cycle of 2 is {2, 4, 8, 6}.
Power cycle of 3 is {3, 9, 7, 1}.
Power cycle of 7 is {7, 9, 3, 1}.
Power cycle of 8 is {8, 4, 2, 6}.