Finding Last Digit of a Number
To find the last digit i.e the unit digit you need to be aware of the cyclicity of the numbers and their powers
2^{1}=2 | 3^{1}=3 | 4^{1}=4 |
2^{2}=4 | 3^{2}=9 | 4^{2}=16 |
2^{3}=8 | 3^{3}=27 | 4^{3}=64 |
2^{4}=16 | 3^{4}=81 | 4^{4}=256 |
2^{5}=32 | 3^{5}=243 | 4^{5}=1024 |
2^{6}=64 | 3^{6}=729 | 4^{6}=4096 |
2^{7}=128 | 3^{7}=2187 | |
2^{8}=256 | 3^{8}=6561 |
We can see that unit's digit of 2^{1},2^{5},2^{9} is 2 and so on. Therefore, after every four powers of 2, the units digit of the number starts repeating.Thus we can say that cyclicity of unit's digit of higher powers of 2 is 4.
Similarily unit digit of power of 4 starts reeating after 2, thus its cyclicity is 2.
Unit digit follows a periodic pattern that is after a particular period it repeats in a cyclic form, this is called cyclicity.
Unit digits of numbers ending with 0,1,5,6 is always the same irrespective of their powers raised on them.
Unit digit of numbers ending with 4,9 follows the pattern with cyclicity of 2
Unit digit of numbers ending with 2,3,7,8 follows the pattern with cyclicity of 4
In order to find the last digit of any number xyz^{abc} whose who's last digit 'z' has the cyclicity of 4, then to find the last digit write the number as z^{4k + m} where 4k+m = abc, and m is smaller than equal to z and not 0.
In order to find the last digit of any number xyz^{abc} whose who's last digit 'z' has the cyclicity of 2, then to find the last digit write the number as z^{2k + m} where 2k+m = abc, and m is smaller than or equal to z ans is not equal to 0.
2 | 3 |
4 | 6 |
2 | 4 |
5 | 6 |
3 | 6 |
1 | 7 |
7 | 1 |
9 | 3 |
1 | 2 |
9 | 0 |