What are the last two digits of 7^(2008)? (1) 21 (2) 61 (3) 01 (4) 41 (5) 81

by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C)

See the cylicity 7^1=07 7^2=49 7^3=43 7^4=01 7^5=07 so on... So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,. Total number of groups are 2008/4=502 Last digit will be 01 suppose if it was 2007 then we will get 3 as remainder and answer will be 7^3=43 Hope it is clear now:)

saurabhmahajan wrote: by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C) sixpointer wrote: So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,. Total number of groups are 2008/4=502 Last digit will be 01 Thanks Guys, I was trying to figure a basic rule with these type of problems but i was not able to figure out one. This is what i got, Its only with 11 We can have the same last two digits with the same powers but the cycle is of 10 over there. Now i understand that last two digits will be same if the number is 7 and 11 did you find any easy to solve a problem if the power if 9^ 2008 or 8 ^ 2008. We can not apply this rule with 6 8 9 ,

goyalsau wrote:saurabhmahajan wrote: by taking 7^2, 7^3, 7^ 4.......7^n we see that the digits 07,49,43,01,07..... are repeated. Out of the answer choices only 01 appears in the above series so IMO: 3 (C) sixpointer wrote: So u can say cycle is repeating after 4 steps . Now to find how many such 4 steps will be there , we consider 4 steps as one group ,. Total number of groups are 2008/4=502 Last digit will be 01 Thanks Guys, I was trying to figure a basic rule with these type of problems but i was not able to figure out one. This is what i got, Its only with 11 We can have the same last two digits with the same powers but the cycle is of 10 over there. Now i understand that last two digits will be same if the number is 7 and 11 did you find any easy to solve a problem if the power if 9^ 2008 or 8 ^ 2008. We can not apply this rule with 6 8 9 , Another method Find the remainder when 7^2008 is divided by 100