last 2 digits.

[goyalsau]

What are the last two digits of 7^(2008)?

(1) 21 (2) 61 (3) 01 (4) 41 (5) 81

[saurabhmahajan]

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]

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:)

[goyalsau]

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)

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 ,

[sixpointer]

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)

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