TIFR CSE 2021 | Part A | Question: 13

Question

What are the last two digits of $7^{2021}$?

• A. $67$
• B. $07$
• C. $27$
• D. $01$
• E. $77$

Answer 1

Option $(B)$

Last two digits are looping as:

$07,49,43,01 | 07,49,43,01 | …..$

 So, $7^{2020}$ will give last two digits as $01$ and $7^{2021}$ will give $07$ as the last two digits.

Answer 2

$7^{2}\mod100=49$

$7^{4}\mod 100=49^{2}\mod 100 = 2401 \mod 100 =1$

$7^{2020}\mod100=(7^{4})^{505}\mod100=1^{505}\mod 100=1$

$7^{2021}\mod100=7.1\mod100=7$

Option (B)

Answer 3

Using Power cycle of 7 which is 4:-
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807.
So Idea is to bring the number to power of 4 as:
   7^2021 = ((7^4)^505) * (7^1) = (2401)^505 * 7

No we care only about the last two digits:-
   (...01)^(anything) = 01 

Now our Final Answer:
           Ans –> 01*7 = 07
Option B.