CYCLICITY
2. Cyclicity of 2: [21 = 2] ; [22 = 4] ; [23 = 8] ; [24 = 16] ; [25 = 32]; ...
At the place of unit digit we will get (2,4,8,6) and again after 4th power this cycle will repeat. So cyclicity of 2 is Four (2,4,8,6).
3. Cyclicity of 3: [31 = 3] ; [32 = 9] ; [33 = 27] ; [34 = 81] ; [35 = 243]; ...
At the place of unit digit we will get (3,9,7,1) and again after 4th power this cycle will repeat. So cyclicity of 3 is Four (3,9,7,1).
4. Cyclicity of 4: [41 = 4] ; [42 = 16] ; [43 = 64] ; [44 = 256] ; ...
At the place of unit digit we will get '4' when the power is odd and '6' when the power is even. So cyclicity of 4 is Two (4,6).
5. Cyclicity of 5: [51 = 5] ; [52 = 25] ; ...
It means we always get 5 at unit place irrespective of its power, so cyclicity of 5 is one (5).
6. Cyclicity of 6: [61 = 6] ; [62 = 36] ; ...
It means we always get 6 at unit place irrespective of its power, so cyclicity of 6 is one (6).
7. Cyclicity of 7: [71 = 7] ; [72 = 49] ; [73 = 343] ; [74 = 2401] ; [75 = 16807]; ...
At the place of unit digit we will get (7,9,3,1) and again after 4th power this cycle will repeat. So cyclicity of 7 is Four (7,9,3,1).
8. Cyclicity of 8: [81 = 8] ; [82 = 64] ; [83 = 512] ; [84 = 4096] ; [85 = 32768]; ...
At the place of unit digit we will get (8,4,2,6) and again after 4th power this cycle will repeat. So cyclicity of 8 is Four (8,4,2,6).
9. Cyclicity of 9: [91 = 9] ; [92 = 81] ; [93 = 729] ; [94 = 6561] ; ...
At the place of unit digit we will get '9' when the power is odd and '1' when the power is even. So cyclicity of 4 is Two (4,6).
Solution: First number is: 13363
Here unit digit is 3 raised to the power 63. We know that cyclicity of 3 is Four (3,9,7,1), and after every fourth power we will get '3' again at unit place.
Step 1: Divide '63' by 4. We will get '3' as remainder. So the third number which is '7' must come at unit place.
⇒ Unit digit of 13363 = 7
In case we get '1' as remainder then unit digit will be '3', if we get '2' as remainder then unit digit will be '9' and if we get '0' as remainder then unit digit will be '1'.
Second number is 439
Cyclicity of 4 is Two (4,6). Here power is odd so unit digit will be 4.
Step 2: Multiply the unit digit of all the numbers.
⇒ 3 x 4 = 12
⇒ Unit digit of 13363 x 439 is '2'.
Example: Find the unit digit of 1111 + 1212 + 1313 + 1414 + 1515
Solution:
It doesn't matter how long a number is, just take unit digits of given numbers, remember their cyclicity, check their powers, divide their powers by cyclicity of respective numbers and determine the unit digits after operation. There is no need to use paper-pen for such questions. Good Luck 👍
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