Concept: Last 2 Digits of a Number


INTRODUCTION

To calculate last two digits of a addition/product of 2 or more numbers, we need to only calculate the addition/product of numbers formed by last 2 digits.

Example: Last 2 digits of 123 + 346 + 987 is same as the last 2 digits of 23 + 46 + 87 which is equal to 56.

Example: Last 2 digits of 123 × 346 is same as the last 2 digits of 23 × 46 which is equal to 58.

 

LAST 2 DIGITS OF POWERS OF ODD NUMBERS ENDING WITH 1, 3, 7 OR 9

Remember:
  • Last 2 digits of (a1)xyz =  Unit's digit of a×z  
    • Units digit of (a1)xyz = 1
    • Tens digit of (a1)xyz = Unit's digit of a × z

The goal is to convert the given number in the form of (a1)power.

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 4th 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 4th 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

 

LAST 2 DIGITS OF POWERS OF NUMBERS ENDING WITH 5

Remember:
  • Last 2 digits of power of a number ending with 5 and whose ten's digit is odd is
    • 25, if the power is even.
    • 75, if the power is odd.
  • Last 2 digits of power of a number ending with 5 and whose ten's digit is even is
    • always 25, for any power.

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 75odd number = 75
⇒ Last 2 digits of 75even 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 45power (> 1) = 25

∴ Last 2 digits of 45250 = 25

 

LAST 2 DIGITS OF POWERS OF EVEN NUMBERS ENDING WITH 2, 4, 6 OR 8

Remember:
  • Last 2 digits of 210 = 24
    • Last 2 digits of 2odd number × 10 = 24
  • Last 2 digits of 220 = 76
    • Last 2 digits of 2even number × 10 = 76
  • Last 2 digits of 220 × 2x is same as last 2 digits of 2x [x > 2]
  • Last 2 digits of (a1)xyz =  Unit's digit of a×z  
    • Units digit of (a1)xyz = 1
    • Tens digit of (a1)xyz = Unit's digit of a × z

To calculate last two digits of power of an even number, we need to first write it in the form of 2a × (odd number)b

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 230 × 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 221 × 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

 

LAST 2 DIGITS OF POWERS OF ANY NUMBERS ENDING WITH 0

Last 2 digits of any power (> 1) of a number ending with 0 will be '00'.

Example: Last 2 digits of (120)276 = 00

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