LAST TWO DIGITS OF A NUMBER

Question:The last two digits of (31^786)(19^266)(2^101) is

A. 14

B. 22

C. 36

D. 72

E. 92

Lets check how to drill this.

We will discuss the last two digits of numbers

ending with the following digits in sets:

a) 1

b) 3,7&9

c) 2, 4,6&8

d) 5

a) Number ending with 1:

Ex : Find the last 2 digits of 31^786

Now, multiply the 10s digit of the number with

the last digit of exponent

31^78

6=

3*6=1

8 ==> 8 is the 10s digit.

Units digit is obviously 1

So, last 2 digits are => 81

b) Number ending with 3 ,7 & 9:

Ex: Find last 2 digits of 19^266

We need to get this in such as way that the

base has last digit as 1

19^266 = (19^2)^133 = 3

61^13

3Now, follow the previous method =>

6 * 3 = 18 = 18

8 is 10s digit.

So, last two digits are => 81

Remember:

3^4=81

7^4=2401

9^2 = 81

Ex 2: Find last two digits of 33^288

Now, 33^288 = (33^4)^72 = (xx

21 )^7

2Tens digit is ->

2*2 = 04 -> 4

So, last two digits are => 41

Ex 3: find last 2 digits of 87^474

(87^2)*(87^4)^118 => (xx69) * (xx

61 )^11

8 ==> (6 x 8 = 48 ==> 8 is 10s digit)

=> (xx69)*(81)

So, last two digits are 89

c)Ending with 2, 4, 6 or 8:

Here, we use the fact that 76 power any

number gives 76. i.e 76^n = ...xxx76 (e.g. 76^2 = 5,776, 76^3 = 438,976)

We also need to remember that,

24^2 = xx76

2^10 xx24

24^even = xx76

24^odd = xx24

Ex: Find the last two digits of 2^543

2^543 = ((2^10)^54) * (2^3)

= ((xx24)54)* 8

= ((xx76)^27)*8

76 power any number is 76

Which gives last digits as => 76 * 8 = 608

So last two digits are : 08

c)Ending with 5:

5 has its own special cases:

The tens digit of a number raised to a power changes based on tens digits of the number itself. Say, we have following number: 05, 15, 25, 35…

say-->

x =

05

x^2 =

25x^3 = 1

25x^4 = 6

25as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15

x^2 = 2

25x^3 = 33

75x^4 = 506

25x^5 = 7593

75as the tens digit of x is 1(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on.

x = 25

x^2 = 6

25x^3 = ..XX

25x^4 = ..XX

25x^5 = ..XX

25as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35

x^2 = 12

25x^3 = ..XX

75x^4 = ..XX

25x^5 = ..XX

75as the tens digit of x is 3(i.e. odd), last two digits repeats after every 2 iterations. It follows pattern of 25,75,25,75...so on

From above, we see that

(1) if number's 10s digit is even, then last digit will always has to be 25.

(2) if number's 10s digit is odd, then last digit will always has to be in cyclic pattern of 25, 75,25,75 and so on...

Now, again, try the question posted on the top.

_________________

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