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

= 6

example : 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 = 5

How 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

= 5

Method 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 = 9

**Finding 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 co-primes, 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 = 3|43, 7^4 = 24|01 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.

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

= 6

example : 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 = 5

How 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

= 5

Method 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 = 9

**Finding 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 co-primes, 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 = 3|43, 7^4 = 24|01 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.