Cyclicity and trailing digits (unit’s digit, ten’s digit and more)  Anubhav Sehgal, NMIMS Mumbai

Finding unit’s digit
Method 1: Find remainder with 10
Unit digit of 68 = 8  which is nothing but 68 mod 10 = 8
Unit digit of 21 * 47 * 18
= Remainder when 21 * 47 * 18 is divided by 10
= 1 * 7 * 8 mod 10
= 6example : Find the unit digit of 45^65.
Unit digit = 45^65 mod 10 = 5^65 mod 10 = 25^32 * 5 mod 10 = 5^33 mod 10
5^33 mod 10 = 25^16 * 5 mod 10 = 5^17 mod 10 = 25^8 * 5 mod 10 = 5^9 mod 10
5^9 mod 10 = 5How do you shorten this process?
Try finding for a larger factor of power on first occasion
45^65 mod 10
= 5^65 mod 10
= 3125^13 mod 10
= 5^13 mod 10
= 3125 * 3125 * 125 mod 10
= 5Method 2: Use Cyclicity of digits
Digit ^1 ^2 ^3 ^4 ^5 ^6 ^7 ^8 Cycle Length 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 2 2 4 8 6 2 4 8 6 4 3 3 9 7 1 3 9 7 1 4 4 4 6 4 6 4 6 4 6 2 5 5 5 5 5 5 5 5 5 1 6 6 6 6 6 6 6 6 6 1 7 7 9 3 1 7 9 3 1 4 8 8 4 2 6 8 4 2 6 4 9 9 1 9 1 9 1 9 1 2 example : Find the unit digit of 37^74
Unit digit of 37^74 = Unit digit of 7^74 = Unit digit of (7^18)^4 * 7^2 = 1 * 9 = 9Finding ten’s digit
Method 1 : Finding remainder by 100
Either directly find remainder by 100 or use Chinese remainder theorem discussed before with 100 = 25 * 4.
Example Find the ten’s place digit of 517^100
Usually it is better to go by Chinese remainder theorem.
517^100 mod 4 = 1^100 mod 4 = 1
517^100 mod 25 = 17^100 mod 25
Since 17 and 25 are coprimes, we may apply Euler’s theorem to reduce the exponent.
E(25) = 20 => 17^20 mod 25 = 1 => (17^20)^5 mod 25 = 1 => 17^100 mod 25 = 1
4a + 1 = 25b + 1
Remainder = 1 or 01
which means ten’s digit = 0 while unit’s digit = 1.Method 2 : Using Cyclicity
Digits Cyclicity 2, 3, 8 20 4, 9 10 5 1 6 5 7 4 So for really high powers you can first reduce them all by factors of 20 since a cycle of 1, 4, 5 or 10 taken multiple times (20, 5, 4, 2 times respectively) is same as a cycle of 20.
Example Find the ten’s place digit of 7^43
Ten’s place digit of 7^43 = Ten’s digit of (7^4)^10 * 7^3 = 01 * 43 = 43.
So, ten’s place digit = 4. [As 7^4 = 2401. So cycle of 7^1 = 07, 7^2 = 49, 7^3 = 343, 7^4 = 2401 repeats itself 10 times.]For further trailing digits like hundredths place digit and so on we take remainders with next powers of 10 like 1000 for hundredths place digit with probably 125 and 8 used for Chinese Remainder Theorem
Patterns and Generalizations for Cyclicity
For odd numbers
 (Any odd number with unit digit 1, 3, 7, 9)^20N will have 01 as the last two digits
example Last two digits of 37^20 = Last two digits of 37^40 = Last two digits of 37^60  (Any odd number with unit digit 5)^N where N >1 will have 25 as the last two digits
 Last two digits are same for 21^5 = 41^5 = 61^5 = 81^5 = 01^5 = 01(last two digits)
11^5 = 31^5 = 51^5 = 71^5 = 91^5 = 51(last two digits)
9^10 = 19^10 = 29^10 = …. = 99^10 = 01(last two digits)  Last two digits for powers of 11 are as follows
11^1 = 11, 11^2 = 21, 11^3 = 31, 11^4 = 41, 11^5 = 51, …. , 11^8 = 81, 11^9 = 91, 11^10 = 01
For even numbers
 (Any even number with unit digit 2, 4, 6, 8)^20N will have 76 as the last two digits.
 (M6)^5N will have last two digits as 76 where M is any digit and power is any multiple of 5.
 (M4)^(odd powers of 5) OR (M8)^(odd powers of 5) will have last two digits as 24.
 (M4)^10N will have last two digits as 76.
 (Any odd number with unit digit 1, 3, 7, 9)^20N will have 01 as the last two digits