Latest Elitmus Aptitude Question SOLUTION: what is the remainder when 30^72^87 is divided by 11. A) 5 B) 7 C) 1 D) None of these

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(#M40153657) ELITMUS Numerical Ability question Keep an EYE

what is the remainder when 30^72^87 is divided by 11.
A) 5
B) 7
C) 1
D) None of these

Asked In Elitmus PRATIM JANA (8 years ago)

Solved Shayak Das Read Solution (20)

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Read Solution (20) :

polo  (8 years ago)

okay here is the general method of how to solve this type of problem

eg. 23^56^78/11 or 63^54^78/6 or x^y^z/(any number) something like that

so lets take the above problem

30^72^87/11 .............. no need to go in detail just follow this method and all problem of this type will be solved.


1. now always divide z or 87 or highest power with 4.

we get 87/4 and take its remainder ,,, basically 87%4=3 and save it .

2. now always divide 56 or y or middle power with 10 .

we get 72/10 and take its remainder ,,,, basically 72%10 = 2

3. now we get z=3 and y=2 ,, now solve this until you get a ( unit digit < 10 )

2^3 or y^z (2^3 = 2 * 2 * 2 / 10 )

4. now divide x or 30 by 11 and get remainder i.e. 30 % 11 = 8

now put the solved y^z or 72^87 = 8 ,,,, here

we get 8 ^ 8 % 11 = 5 -----> (8 * 8 * 8 * ............... 8 times % 11 )

use this for all the problems of this type divide y by 10 and z by 4.

and quickly get the answer .

Is this solution Helpfull? Yes (40) | No (6)  

Shayak Das  (8 years ago) Best Solution

Ans is 5 option (a)

Is this solution Helpfull? Yes (16) | No (7)  

Niraj Kumar  (8 years ago)

let us take N=72^87
eq is..30^N/11=8^N/11
here cyclicity of 8 is 10
then N/10=72^87/10=2^87/10
here cyclicity of 2 is
then 87%4=3(3rd cycle of 2)....i.e2^3%10=8(8th cycle of8)
therefore final ans is 8^8%11=5

Is this solution Helpfull? Yes (14) | No (3)  

NARESH MAJUMDER  (8 years ago)

Rem [30^72^87 / 11] = Rem [(-3)^72^87 / 11] = Rem [3^72^87 / 11]

Now, we need to observe the pattern
3^1 when divided by 11, leaves a remainder of 3
3^2 when divided by 11, leaves a remainder of 9
3^3 when divided by 11, leaves a remainder of 5
3^4 when divided by 11, leaves a remainder of 4
3^5 when divided by 11, leaves a remainder of 1
And then the same cycle of 3, 9, 5, 4 and 1 will continue.
If a number is of the format of 3^(5k + 1), it will leave a remainder of 3
If a number is of the format of 3^(5k + 2), it will leave a remainder of 9
If a number is of the format of 3^(5k + 3), it will leave a remainder of 5
If a number is of the format of 3^(5k + 4), it will leave a remainder of 4
If a number is of the format of 3^(5k), it will leave a remainder of 1

The number given to us is 3^72^87
Let us find out Rem[Power / Cyclicity] t0 find out if it 3^(5k + what?)

Rem [72^87 / 5]
= Rem [2^87 / 5]
= Rem [2*4^43/5]
= Rem [2*(-1) / 5]
= -2
= 3
=> The number is of the format 3^(5k + 3)
=> Rem [3^72^87 / 11] = 5

Is this solution Helpfull? Yes (8) | No (1)  

Priyesh jain  (8 years ago)

i think it will be none of these...
as 30/11 will give us reminder of 8
now cyclicity of 8 when divided by 11 is 10 as
8^1%11=8;
8^2%11=9;
8^3%11=(break)8^2*8^1=(as previous reminder)8*9=72%11=6;
8^4%11=4;
8^5%11=10;
8^6%11=3;
8^7%11=2;
8^8%11=5;
8^9%11=7;
8^10%11=1;
Now as the rule of powers 72*87=6264;
we can break
((8^10)626)%11*(8^4)%11= (1)^626 * 4096%11=1*4
so ANS will be 4....
so ans is d...
PLZ
Notify me if any mistake....

Is this solution Helpfull? Yes (5) | No (4)  

akshay kumar  (8 years ago)

30^72^87/11

in this type of question u have get value 1^something then u can give the answer easily
if u know cyclicity then u can get answer easily.
first 30/11=8 reminder
now this will become 8^72^87/11
find the reminder of 8/11 if u get 1 then u came to answer
8/11=3
8^2=8*8/11=2
8^3=5 ------(i)
8^4=7
8^5=1 here we get 1 ok
now here cyclicity is 5
so we have to dive 72^87/5= 2 as reminder then
now cycle of 2 is 4 so 87/4=3 reminder now go to the ---(i)
where u get the answer .... as 5.

Is this solution Helpfull? Yes (4) | No (4)  

Abhishek  (8 years ago)

ok i have read all the solutions here
some people were asking how to get 8^8/11 remainder?
it is simple
=(11-3)^8/11
=(-3)^8/11
=3^8/11
=3^4*3^4/11
=81*81/11
=(77+4)(77+4)/11
since 77 is divided by 11 so again it will become
=4*4/11
=16/11
now you know remainder would be 5

Is this solution Helpfull? Yes (4) | No   

ARNAB PAL  (8 years ago)

c) 1 is the ans.

Is this solution Helpfull? Yes (3) | No (3)  

Abhijit Das  (8 years ago)

72^87 can be written as 10k+8
so 30^(10k+8)/11
=30^(11-1)^k . 30^8/11
now from the formula a^(p-1)/p gives a remainder 1 where p is a prime number.
so 30^(11-1)^k gives remainder 1.
now 30^8/11 gives remainder 5.( go through remainder theorem. you will find)
so remainder is.... 5

Is this solution Helpfull? Yes (3) | No (1)  

Anjali Sadhukhan  (8 years ago)

hi guys,i dont get how 8^8%11 gives the remainder as 5.i would be thankful if any1 spare a few time in explaining it.

Is this solution Helpfull? Yes (3) | No (1)  

Asmita  (8 years ago)

sry m not able to undrstand any solutn .. plz suggest me in brief :(

Is this solution Helpfull? Yes (3) | No (2)  

Anuj Awasthi  (8 years ago)

polo bhai agar %4 lene me 0 ajaye to is question me kya krenge

(50 ki power 56 ki power 52)%11

Is this solution Helpfull? Yes (2) | No   

suraj maurya  (8 years ago)

Ans is 4 hence 4 is not there hence Ans is(D)None of this

Is this solution Helpfull? Yes (2) | No (2)  

sonkar  (8 years ago)

sol:
30%11=8 72%11=6 87%11=10
now 8*6*0=480%11=7
so ans:7

Is this solution Helpfull? Yes (1) | No   

yamini  (8 years ago)

plz exaplain it

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shikher  (8 years ago)

Please explain the solution

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Neeraj Shukla  (8 years ago)

correct ans is 7
devide each no individual like (30%11= -3) and (72%11= 6) and (87%11= -1)
now multiply each remender and devide again get 7 remender

Is this solution Helpfull? Yes | No (3)  

Bushra  (8 years ago)

please somebody explain me how to find 8^8%11=5.............. ill be very thankful

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kumari anjana  (8 years ago)

sory
i can not

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Manas Pant  (4 years ago)

For those who wanted solution for 8^8%11=5
8=2^3
therefore we can write
(2^3)^8 which is
2^24
now 2 repeats cycle after 4 i.e.
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32...repeats

therefore 2^24 = 2^4=16
now
16%11=5
Hence the answer

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