0 votes 0 votes What are the last two digits of $7^{2008}?$ $21$ $61$ $01$ $41$ $81$ Quantitative Aptitude cat2008 quantitative-aptitude number-systems + – go_editor asked Nov 26, 2015 • edited Mar 28, 2022 by Lakshman Bhaiya go_editor 13.8k points 939 views answer comment Share See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Powers of 7 from 0 are as follows: 1 7 49 343 2401 ... - so the cycle is repeating after every 4 powers. $7^{2008} = 7^{4 \times 502}$. So, the last two digits must be "01". Arjun answered Nov 28, 2015 • selected Dec 5, 2015 by Pooja Palod Arjun 8.6k points comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes answer c . That is 01. $7^{2008}$ equals can be written as $7^{4*502}$ now $7^4$ equals $49*49$ whose last 2 digits are 01 and then whenever a number has 01 as power something. take the tens place and multiply by the last digit of power and that thats number unit digit and consider it as its tens digit. ( case 1) so here unit digit will be 1 and we have to find tens place. so take the tens place of the found last 2 digits 01 which is 0 . multiply with 2=0. so tens place will be also zero . so 01. here are all the case of such a problem . http://www.campusgate.co.in/2011/10/finding-unit-and-last-two-digits-of.html Tendua answered Nov 28, 2015 Tendua 188 points comment Share See all 0 reply Please log in or register to add a comment.