Concept: Last 2 Digits of a Number

Example: Find the last 2 digits of (131)⁹⁹

Solution:

Last 2 digits of (131)⁹⁹ = Last 2 digits of (31)⁹⁹

Unit's digit of (131)⁹⁹ = 1

Ten's digit of (131)⁹⁹ = Unit's digit of 3 × 9 = 7

Last 2 digits of (131)⁹⁹ = 71

Example: Find the last 2 digits of (133)⁹⁹

Solution:
Last 2 digits of (133)⁹⁹ = Last 2 digits of (33)⁹⁹

We need to convert the given number in the form of (a1)^power

We know 4^th power of 3 gives last digit as 1.

Last 2 digits of (33)⁹⁹ = Last 2 digits of (33)⁹⁶ × 33³ 

= Last 2 digits of ((33)⁴)²⁴ × 33³ = Last 2 digits of ((89)²)²⁴ × 33² × 33

= Last 2 digits of (21)²⁴ × 89 × 33

= Last 2 digits of 81 × 89 × 33

= 97

Example: Find the last 2 digits of (257)¹³³

Solution:
Last 2 digits of (257)¹³³ = Last 2 digits of (57)¹³³

We need to convert the given number in the form of (a1)power

We know 4^th power of 7 gives last digit as 1.

Last 2 digits of (57)¹³³ = Last 2 digits of (57)¹³² × 57 

= Last 2 digits of ((57)⁴)³³ × 57 = Last 2 digits of ((49)²)³³ × 57

= Last 2 digits of (01)³³ × 57

= Last 2 digits of 01 × 57

= 57

Unit's digit of (01)³³ = 1
Ten's digit of (01)³³ = Unit's digit of 0 × 3 = 0
Last 2 digits of (01)³³ = 01

Example: Find the last 2 digits of (569)¹⁷⁰

Solution:
Last 2 digits of (569)¹⁷⁰ = Last 2 digits of (69)¹⁷⁰

We need to convert the given number in the form of (a1)^power

We know 2nd power of 9 gives last digit as 1.

Last 2 digits of (69)¹⁷⁰ = Last 2 digits of ((69)²)⁸⁵  

= Last 2 digits of (61)⁸⁵

= 01

Unit's digit of (61)⁸⁵ = 1
Ten's digit of (61)⁸⁵ = Unit's digit of 6 × 5 = 0
Last 2 digits of (61)⁸⁵ = 01

Example: Find the last 2 digits of (475)⁵⁸⁹

Solution:
Last 2 digits of (475)⁵⁸⁹ = Last 2 digits of (75)⁵⁸⁹

Last 2 digits of 75¹ = 75
Last 2 digits of 75² = 25
Last 2 digits of 75³ = 75
Last 2 digits of 75⁴ = 25
Last 2 digits of 75⁵ = 75
Last 2 digits of 75⁶ = 25

⇒ Last 2 digits of 75^{odd number} = 75
⇒ Last 2 digits of 75^{even number} = 25

∴ Since 589 is odd, Last 2 digits of 75⁵⁸⁹ = 75

Example: Find the last 2 digits of (345)²⁵⁰

Solution:
Last 2 digits of (345)²⁵⁰ = Last 2 digits of (45)²⁵⁰

Last 2 digits of 45¹ = 45
Last 2 digits of 45² = 25
Last 2 digits of 45³ = 25
Last 2 digits of 45⁴ = 25

⇒ Last 2 digits of 45^{power (> 1)} = 25

∴ Last 2 digits of 45²⁵⁰ = 25

Example: Find the last 2 digits of (472)¹⁰⁰

Solution:
Last 2 digits of (472)¹⁰⁰ = Last 2 digits of (72)¹⁰⁰

We need to convert the given number in the form of (2)^a × (odd)^b

∴ 72 can be written as 2³ × 9
∴ 72¹⁰⁰ can be written as 2³⁰⁰ × 9¹⁰⁰

Last 2 digits of 2³⁰⁰ = Last 2 digits of 2^{30 × 10} = 76

Last 2 digits of 9¹⁰⁰ = Last 2 digits of 81⁵⁰ = 01

⇒ Last 2 digits of 72¹⁰⁰ = Last 2 digits of 76 × 01 = 76

Example: Find the last 2 digits of (54)²¹⁰

Solution:

We need to convert the given number in the form of (2)^a × (odd)^b

∴ 54 can be written as 2 × 27
∴ 54²¹⁰ can be written as 2²¹⁰ × 27²¹⁰

Last 2 digits of 2²¹⁰ = Last 2 digits of 2^{21 × 10} = 24

Last 2 digits of 27²¹⁰ = Last 2 digits of (27²)¹⁰⁵
Last 2 digits of 27²¹⁰ = Last 2 digits of (29)¹⁰⁵
Last 2 digits of 27²¹⁰ = Last 2 digits of (29²)⁵² × 29
Last 2 digits of 27²¹⁰ = Last 2 digits of (41)⁵² × 29
Last 2 digits of 27²¹⁰ = Last 2 digits of 81 × 29
Last 2 digits of 27²¹⁰ = 49

⇒ Last 2 digits of 54²¹⁰ = Last 2 digits of 24 × 49 = 76