when we need to find the last digit ...only consider the last digit of both base and power...
here 27 last digit is 7...and 36 last digit is 6...
so u need to jus consider 7 ^ 6.
which now boils down to power cycle theorem....and power cycle of 7 is 4 ....so now we are left with 7^2 therefore last digit is 9..:)...
hope this help u 😃
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
27^36
Consider the last digit here it's 7...
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
___________
7^5 = 7
So actually the cyclicity of 7 is 4....
Now with the problem :
27^36
Ignore the ten's digit and proceed with the units digit ( Coz U r asked to find the unit's digit )
7^36
Can b written as
7^4 * 7^4......... ( 9times )
Value of 7^4 = 1 (We derived it )
So value of 7^4 * 7^4......... ( 9times ) is 1 * 1* 1 ( 9 times ) = 1....
there is an easy method to this que as given in arun sharma.
27^36 can be written as 729^18. now 729 when divided by 7 gives the remainder 1. now 1^18 gives 1 which is the answer. the approach used by abhishek is also fyn but will take 1 extra minute to solve coz it involves a bit of analysis. the approach i have used can be used in any of the question of this type.
for eg we have 25^102 and we have to find remainder when divided by 17. we proceed by 225^51.dividing 225 by 17 gives 4. now its 4^51.now u have to think a bit. if we take one four out it reduces to 4*(4^50).now 4^5 is 256 which gives remainder 1 when divided by 17. the problem now reduces to 4*(256^10) which gives 4 as remainder.
it seems a long process but actual execution takes half a min. after practice and it wont take even half a min.
when we need to find the last digit ...only consider the last digit of both base and power...
here 27 last digit is 7...and 36 last digit is 6...
so u need to jus consider 7 ^ 6.
which now boils down to power cycle theorem....and power cycle of 7 is 4 ....so now we are left with 7^2 therefore last digit is 9..:)...
hope this help u :)
Sorry dude, but for the power, u need not consider the last digit. U need to take care of the entire exponent.
In this case, 27^36, we need to find 7^36.
Cyclicity of 7 is 4, and 36 % 4 = 0. So, 7^6 = 7^4 = 1 (just considering the last digit). Answer is 1, not 9.
thanks fr correcting me:)
PLZ answer this:
The average of three consecutive prime nos. is 223/3 . What is the difference between the greatest and the smallest number that can be part of such a set? a) 8 b) 14 c) 16 d) 10
According to me the nos are 71,73,79.....so it will be a)8.....BUT the ans given in book is 16. PLZ explain why??
PLZ answer this:
The average of three consecutive prime nos. is 223/3 . What is the difference between the greatest and the smallest number that can be part of such a set? a) 8 b) 14 c) 16 d) 10
According to me the nos are 71,73,79.....so it will be a)8.....BUT the ans given in book is 16. PLZ explain why??
67, 73 and 83 !
anonymous:) Says67, 73 and 83 !
but it is written as three consecutive prime nos....then how is it possible..plz explain
SourabhBhaduri Saysbut it is written as three consecutive prime nos....then how is it possible..plz explain
JEEEEE
dint saw the consecutive thingi 
Even i dont think that there exist any such consecutive prime trio :splat::splat:
Ans will be (a)8 (71+73+79)/3 so 79-71 = 8 no other set of prime no's can have an average of 223/3
Me too getting the same answer: 8.
PLZ answer this:
The average of three consecutive prime nos. is 223/3 . What is the difference between the greatest and the smallest number that can be part of such a set? a) 8 b) 14 c) 16 d) 10
According to me the nos are 71,73,79.....so it will be a)8.....BUT the ans given in book is 16. PLZ explain why??
Me too getting 8 :splat:.
since..numbers should be lying around 70 number..and there are so many primes around 70
:shocked:
Please Help..
If 11 sweets are distributed among four boys..then which one is true..
A.two boy recived more than 1 sweet
B.One of boys recvd more than 3 sweets.
C.One of boys recieved fewer than 3 sweets.
D.One of boy rcvd exactly 2 sweets.Iam getting all as correct:splat::splat::splat:
but answer is something else..help
ans shud be all a,b,c,d
1.7+2+1+1 = 11 (2 boys more than one)
2.7+2+1+1 = 11 (1 boy more than 3)
3.4+3+3+1 = 11 (1 boy fewer than 3)
4.7+2+1+1 = 11 (1 boy exactly 2 )
Please Help..
If 11 sweets are distributed among four boys..then which one is true..
A.two boy recived more than 1 sweet
B.One of boys recvd more than 3 sweets.
C.One of boys recieved fewer than 3 sweets.
D.One of boy rcvd exactly 2 sweets.Iam getting all as correct:splat::splat::splat:
but answer is something else..help
bro is the answer 'C'
If yes, i will explain my logic.
bro is the answer 'C'
If yes, i will explain my logic.
yessssss..
Please Help..
If 11 sweets are distributed among four boys..then which one is true..
A.two boy recived more than 1 sweet
B.One of boys recvd more than 3 sweets.
C.One of boys recieved fewer than 3 sweets.
D.One of boy rcvd exactly 2 sweets.
but answer is something else..help
if we consider the events occur independently, then all are correct...
for simultaneous occurrence, let us consider a case of dividing 11 into 1-1-7-2
condition A : 1-1-7-2
condition B : 1-1-7-2
condition C : 1-1-7-2(here it is not said that exactly one receive condition D : 1-1-7-2
so here also we are getting all correct...
ps: is the provided data sufficient? or it seems that I am missing something :lookround:
EDIT: got it...I was not considering the "necessarily true" condition...going to the other way round...
Please Help..
If 11 sweets are distributed among four boys..then which one is true..
A.two boy recived more than 1 sweet
B.One of boys recvd more than 3 sweets.
C.One of boys recieved fewer than 3 sweets.
D.One of boy rcvd exactly 2 sweets.
but answer is something else..help
see, whenever these type of statement is given, you have to find the ''necessarily true'' condition...i.e approach the reverse way you did :)
1st case-- you can easily have 2 boys with only 1 choco each and other two with rest 9.
2nd case-- not necessary again---it can be 3-3-3-2
3rd case--have to be necessary
4th case-- not necessary again--it can be 4-3-3-1 or many such combinations..
hope it helps 😃
if we consider the events occur independently, then all are correct...
for simultaneous occurrence, let us consider a case of dividing 11 into 1-1-7-2
condition A : 1-1-7-2
condition B : 1-1-7-2
condition C : 1-1-7-2(here it is not said that exactly one receive condition D : 1-1-7-2
so here also we are getting all correct...
ps: is the provided data sufficient? or it seems that I am missing something :lookround:
Answer given is C..and i have no idea about it..
I have to contact abhisek bacchan
see, whenever these type of statement is given, you have to find the ''necessarily true'' condition...i.e approach the reverse way you did :)
1st case-- you can easily have 2 boys with only 1 choco each and other two with rest 9.
2nd case-- not necessary again---it can be 3-3-3-2
3rd case--have to be necessary
4th case-- not necessary again--it can be 4-3-3-1 or many such combinations..
hope it helps :)
Dude, if you want to prove option 'a' as false, it should be 1-1-1-8.
As per you, it satisfies the 1st condition, that is it makes it true.
Still, the answer is c. Just the reasoning for the first option u gave seems incorrect.
see, whenever these type of statement is given, you have to find the ''necessarily true'' condition...i.e approach the reverse way you did :)
1st case-- you can easily have 2 boys with only 1 choco each and other two with rest 9.
2nd case-- not necessary again---it can be 3-3-3-2
3rd case--have to be necessary
4th case-- not necessary again--it can be 4-3-3-1 or many such combinations..
hope it helps :)
one small confusion..
one of boys means we can take more than one boy or exactly one boy??..
thanks for answer