1 votes 1 votes What are the last two digits of $7^{2021}$? $67$ $07$ $27$ $01$ $77$ Quantitative Aptitude tifr2021 quantitative-aptitude modular-arithmetic + – soujanyareddy13 asked Mar 25, 2021 retagged Nov 24, 2022 by Lakshman Bhaiya soujanyareddy13 494 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 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. jatinmittal199510 answered Mar 25, 2021 jatinmittal199510 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $7^{2}\mod100=49$ $7^{4}\mod 100=49^{2}\mod 100 = 2401 \mod 100 =1$ $7^{2020}\mod100=(7^{4})^{505}\mod100=1^{505}\mod 100=1$ $7^{2021}\mod100=7.1\mod100=7$ Option (B) vishnu_m7 answered Mar 25, 2021 vishnu_m7 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 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. Yashvir answered Apr 10, 2021 Yashvir comment Share Follow See all 0 reply Please log in or register to add a comment.