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

Question

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

D

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
 3 1^78 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 6 1^13 3 
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 = (xx 2 1 )^7 2 
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) * (xx 6 1 )^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 =  0 5
x^2 =  25 
x^3 = 1 25 
x^4 = 6 25 
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15
x^2 = 2 25 
x^3 = 33 75 
x^4 = 506 25 
x^5 = 7593 75 
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 = 6 25 
x^3 = ..XX 25 
x^4 = ..XX 25 
x^5 = ..XX 25 
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35
x^2 = 12 25 
x^3 = ..XX 75 
x^4 = ..XX 25 
x^5 = ..XX 75 
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.

rajathpanta · Answer

Very laborious isnt it?

goutamread · Answer

Question:The last two digits of (31^786)(19^266)(2^101) is
A. 14
B. 22
C. 36
D. 72
E. 92

Solution: 
31^786==>
Unit Digits: as last digit is 1, units digit has to be 1.
Tens digit:  3 1^78 6  ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8  and thus, 8 is the tens digit.
Thus, last two digit of expression 31^786 = 81

19^266 ==>
Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1.
Tens digit: 19^266 = (19^2)^133 = (xx 6 1)^13 3  ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8  and thus, 8 is the tens digit.
Thus, last two digit of expression 19^266 = 81

(2^101) ==>
Tens Digit: 2^101 = 2^100 * 2 = *2 =  *2 ==> As we know xx24^even = xx76 ==>  *2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52

THUS XX81 * XX81 *XX52 = XX 72

goutamread · Answer

Rajath - This example was just to elaborate all possible scenarios i.e. (a)1 (b)3,7,9 and (c)2,4,6,8. Indeed, I have detailed each step over here. Once you are used to the concept/rules, this problem shall not take more than the (per question) average time.

MunniB · Answer

Interestin methodology. Thanks for the solution

chiknichameleon · Answer

Question:The last two digits of (31^786)(19^266)(2^101) is
A. 14
B. 22
C. 36
D. 72
E. 92

Solution: 
31^786==>
Unit Digits: as last digit is 1, units digit has to be 1.
Tens digit:  3 1^78 6  ==> as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8  and thus, 8 is the tens digit.
Thus, last two digit of expression 31^786 = 81

19^266 ==>
Unit Digits: as last digit is 9, as we know pattern power raise to 9: 9,1,9,1... i.e. it repeats after every 2 iterations.. as 266 is even i.e. dividible by 2, we cud say,units digit has to be 1.
Tens digit: 19^266 = (19^2)^133 = (xx 6 1)^13 3  ==> now, as last digit is 1, we cud directly multiply 3 and 6 ==> 3*6=1 8  and thus, 8 is the tens digit.
Thus, last two digit of expression 19^266 = 81

(2^101) ==>
Tens Digit: 2^101 = 2^100 * 2 = *2 =  *2 ==> As we know xx24^even = xx76 ==>  *2 ==> as we knoe anything raised to xx76 is xx76 ==> xx76*2 = XX52

THUS XX81 * XX81 *XX52 = XX 72 [/quotew

what would be second last digit if last digit is 5

goutamread · Answer

Actually, 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 =  0 5
x^2 =  25 
x^3 = 1 25 
x^4 = 6 25 
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 15
x^2 = 2 25 
x^3 = 33 75 
x^4 = 506 25 
x^5 = 7593 75 
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 = 6 25 
x^3 = ..XX 25 
x^4 = ..XX 25 
x^5 = ..XX 25 
as the tens digit of x is 0(i.e. even), last two digits has to be 25.

x = 35
x^2 = 12 25 
x^3 = ..XX 75 
x^4 = ..XX 25 
x^5 = ..XX 75 
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...

P.S. Thanks for asking the question. I missed that part. I shall update the post to include that part.

chiccufrazer1 · Answer

but during the gmat exam we are always against time..can that question come on gmat exam or maybe there is another way of answering the question?

Posted from my mobile device

Hapreet Singh · Answer

How will we solve this question - 
Q. What will be the last two digits in 1122^1122! ?
(asked in TCS placement Paper).

mcwoodhill · Answer

this is really a good way. my way can lead to the answer but definitely takes way longer time. Thanks a lot, bro.

Mathivanan Palraj · Answer

The answer is 72
31^786 = (30+1)^786 use binomial theorem
The last two digits will be 786*30 + 1 = 81
19^266 = 61^133 
133*60+1 = 81
2^101=(2^10)^10*2=24^10*2=76^5*2=76*2=52(the last two digits of any power of 76 is always 76)
81*81*52=61*52=72

bumpbot · Answer

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

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