shashank.ism wrote:What are the last two digits of 72008?

a) 21
b) 61
c) 01
d) 41
e) 81

Apparently it is 08...:)...Did you miss anything over here shashank??

shashank.ism wrote:What are the last two digits of 72008?

a) 21
b) 61
c) 01
d) 41
e) 81

Thats not fun; what are the last two digits of 7^2008 would have been fun {which requires modulo operations and I think is out of scope again}

ajith wrote:

shashank.ism wrote:What are the last two digits of 72008?

a) 21
b) 61
c) 01
d) 41
e) 81

Thats not fun; what are the last two digits of 7^2008 would have been fun {which requires modulo operations and I think is out of scope again}

Well,this is also okay.
Now,7^1 = 07 [ last 2 digits are written on the RHS of =]
7^2 = 49
7^3 = 43(343)
7^4= 01 (2401)
7^5 = 07
So,we see that after every 4 no.s the same last 2 digits are repeated.[spoiler]
As, 2008 is divisible by 4 so the abs. is 01. (C)[/spoiler]

Ajith,if you carefully note,rest of the 4 options are bogus.If any option would have been 49 or 43 or 07 then the question would have been difficult.

harsh.champ wrote: Ajith,if you carefully note,rest of the 4 options are bogus.If any option would have been 49 or 43 or 07 then the question would have been difficult.

Well you cant say that, it is easily found out that last digit of 7^2008 is 1 so all the options have 1 as the last digit. It would have been easier had they kept the options as u have mentioned

harsh.champ i got as far as 7^4 then i'm like.. yeah i'm doing something wrong... not thinking to myself "just see if it begins repeating somewhere." i wouldn't have known to then divide 2008 by 4, to get to 01 being the answer. I will now in the future though