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

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
31^786=3*6=18 ==> 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 = 361^133
Now, 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 = (xx21 )^72
Tens 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) * (xx61 )^118 ==> (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 = 25
x^3 = 125
x^4 = 625
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15
x^2 = 225
x^3 = 3375
x^4 = 50625
x^5 = 759375
as 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 = 625
x^3 = ..XX25
x^4 = ..XX25
x^5 = ..XX25
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35
x^2 = 1225
x^3 = ..XX75
x^4 = ..XX25
x^5 = ..XX75
as 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.