Latest TCS Aptitude Question SOLUTION: 30^72^87 divided by 11 gives remainder

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(#M40021865) TCS Numerical Ability question Keep an EYE

30^72^87 divided by 11 gives remainder

Asked In TCS Kunal Verma (10 years ago)

Unsolved Read Solution (22)

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VIKAS SINGH  (10 years ago)

72^87 = 10k + 8
=>30^72^87 = 30(10k + 8) = 30^8mod11 = (-3)^8mod11 = 3^3mod11 [ 3^5 = 1mod11]
= 27mod11 = 5mod11.
so remainder is 5.

Is this solution Helpfull? Yes (35) | No (21)  

sivaraman  (10 years ago)

30^72^87=30^6264/11
=(33-3)^6264/11
using binomial theorem,
the last term in the expansion is (-3)^6264/11
=((-3)^2)^3132/11
=9^3132/11
=(11-2)^3132/11
the last term in the expansion is
=(2^(5*626)+2)/11=(32^626)*4/11
((33-1)^626)*4/11=4/11 remainder is 4

Is this solution Helpfull? Yes (34) | No (39)  

Shreyansh Jain  (10 years ago)

5 will be the correct answer
7287 = 10k + 8
=> 3072^87 = 30(10k + 8) = 308mod11 = (-3)8mod11 = 33mod11 [ 35 = 1mod11]
= 27mod11 = 5

Is this solution Helpfull? Yes (8) | No (2)  

sameer singh  (10 years ago)

ans is 1

Is this solution Helpfull? Yes (7) | No (18)  

Tanu  (10 years ago)

Actually what i did wrong is 5*4*9=180
Now, 180/11 will give remainder 4.
So 4 is the correct answer.

Is this solution Helpfull? Yes (5) | No (17)  

RAVI GANGWAR  (10 years ago)

ans=5
sollution:
72^87=10k+a
a=find the unit digit of this number=8
means 72^87=10k+8
30^(10K+8) where 30^10k remainder is 0
then we take 30^8mod11=(-3)^8mod11=(3)^5*3^3mod11=1*3^3Mod11=5 ans

Is this solution Helpfull? Yes (5) | No   

Arti Pawar  (7 years ago)

Here the base is 30 and its exponent is 72^87.

Find out the units digit of exponent i.e of 72^87 which will be 8.
The problem is now reduced to 30^8.
Expand the function in terms of factor of 11; (22 + 8)^8.
In binomial expansion, all the term will be divisible by 11 except 8^8.
Again expand it as 64^4 = (66 - 2)^4.
Here also all the term will be divisible by 11 except (-2)^4 = 16.

The final remainder will be 5

Is this solution Helpfull? Yes (5) | No (1)  

ravi jha  (10 years ago)

For 3 multiples
3 9 27 81 243 729

Remainder when devided by 11:- 3 9 5 4 1 RESP

For 2,
2 4 8 16 32 64 128 256 512 1024 2048 4096


Remainder when devided by 11:2 4 8 5 10 9 7 3 6 1

For 5 multiples
5 25 125 625 3125

Remainder when devided by 11: 5 3 4 9 1

5^5 WILL give reminder 1
2^10 WILL GIVE REMINDER 1
3^5 WILL GIVE REMINDER 1


30^72^87/11=(2.3.5)72^87= 2^72^87.3^72^87.5^72^87

2^6264.3^6264.5^6264/11=2^6260.2^4.3^6260.3^4.5^6260.5^4/11=2^4.3^4.5^4/11=2^4.15^4/11=2^4.4^4/11=2^12/11=4096/11
ans=4

Is this solution Helpfull? Yes (4) | No (3)  

Tanu  (10 years ago)

7 is the answer.

It can b written as (2*3*5)^6264
= 2^6264 *3^6264 * 5^6264
Now when 2^4 divided by 11 the remainder is 5
when 2^14 divided by 11 the remainder is 5
when 2^24 divided by 11 the remainder is 5
when 2^34 divided by 11 the remainder is 5
and so on....

Same u can do for 3 and 5

so ((2^6264)/11) , remainder will be 5
((3^6264)/11) , remainder will be 4
((5^6264)/11) , remainder will be 9

So total 5+4+9=18
Now 18/9 will give remainder 7.

Is this solution Helpfull? Yes (3) | No (28)  

ravi jha  (10 years ago)

2multiples:2 4 8 16 32 64 128 256 512 1024 2048 4096
Remainder when devided by 11:2 4 8 5 10 9 7 3 6 1 resp.

So, 2^10 gives remainder as 1.


30^72^87/11=(22+8)^6264/11 Remainder will be same as 8^6264/11

8^6264/11= 2^(3*6264)=2^18792/11= 2^18790.2^2/11=2^2/11=4/11

So,Remainder=4

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

vinay H M  (10 years ago)

vikas..plzzz explain me in detail

Is this solution Helpfull? Yes (1) | No (2)  

subhan kumar  (10 years ago)

it is in d form a^m^n ... it is not in d form (a^m)^n = a^(m*n) .... how can you multiply 72 and 87 !!

Is this solution Helpfull? Yes (1) | No   

divyang  (10 years ago)

http://totalgadha.com/mod/forum/discuss.php?d=2594

see it 4 this type of que

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Namrata Singh  (10 years ago)

6. 30^72^87 divided by 11 gives remainder

Is this solution Helpfull? Yes (1) | No   

Gaurav Singh  (9 years ago)

7 should be the answer

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raviteja  (9 years ago)

3010 or 810when divided by 11 remainder is 1.
The unit digit of 7287 is 8 (using cyclicity of unit digits) Click here
So 7287 = 10K + 8
30(10K+8)11=(3010)K.30811=1k.30811
8811=22411=(25)4.2411=1611=5

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akshat sinha  (8 years ago)

This should take some thought. Let's start with 30 and work our way up. When you divide 30 by 11, you get a quotient of 2 and a remainder of 8.

How about 302 ? It's 30 times 30. Each 30 divided by 11 gives a remainder of 8. So

30⋅30=(2⋅11+8)(2⋅11+8)=∗⋅11+82

where * is some number we don't care about. And 82=64 gives a remainder of 9 when divided by 11. Aha! The remainder for 302 is the same as the remainder for 82.

Now we can go back to the earlier question. 87287 when divided by 11 gives the same as 88 when divided by 11, and according to our first table, that remainder was 5.

Therefore, we have the answer to the original question. The remainder of 307287 when divided by 11 is 5.

As I was not able to post the pic of table
Please click here..
http://linkshrink.net/7uQbvI
I am sure you get solution so easily..

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kanishk kalra  (9 years ago)

72^87 = 10k + 8
=> 3072^87 = 30(10k + 8) = 308mod11 = (-3)8mod11 = 33mod11 [ 35 = 1mod11]
= 27mod11 = 5mod11.

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Arsha hasan  (8 years ago)

ans is 1
(30^72^78)/11
(30^6264)/11
((3^5)^6259)/11
(243/11)^6264 ---> R--> ((1)^6264)/11
one raise to the power any number is one.

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Rakesh Sharma  (8 years ago)

----- BY using fermets theorem
a^(p-1)=1modp
so , write the given qst as 30^6264
p=11,a=30

30^10=1mod11
30^4 ((30^10)^626)=1^6260* 30^10 mod 11
so
30^10 mod 11=4
so ans is 4.

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shubham  (8 years ago)

Fermat little theorem says, (ap−1)/p remainder is 1. ie., 3010 or 810when divided by 11 remainder is 1.
The unit digit of 7287 is 8 (using cyclicity of unit digits) So 7287 = 10K + 8
30(10K+8)/11=(3010)K.308/11=1k.308/11
8^8/11=2^24/11=(25)4.24/11=16/11=5(we can write (2^5)^4 also as (2^10)^2/11=1) or
then we take 30^8mod11=(-3)^8mod11=(3)^5*3^3mod11=1*3^3Mod11=5 ans. (i.e. either take 30-22=8 or 30-33=-3)

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Kavinkumar  (4 years ago)

ans=5
because so many persons were solved the solution...
then i also agree these type of answer

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