Latest TCS Aptitude Question SOLUTION: what is the remainder of (128)^1000 %9 ?

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(#M40029610) TCS Logical Reasoning question Keep an EYE

what is the remainder of (128)^1000 %9 ?

Asked In TCS shreya (10 years ago)

Unsolved Read Solution (23)

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Read Solution (23) :

Md Yeanur Mallick  (10 years ago)

(128)^1000/9=(126+2)^1000
==>(2^3)^111 *2
==>(9-1)^111 *2
==> -(1)^111 *2
==>-2
so 7 is the remainder

Is this solution Helpfull? Yes (22) | No (6)  

Dipra Dutta  (10 years ago)

128^1000/9 = 2^1000/9 = (8)^333*2/9= 8*2/9 = 16/9 = 7 will be the remainder. tats it.

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Athena Chakraborty  (10 years ago)

Guys..pls check this site...
http://www.pagalguy.com/news/cat-2012-quantitative-aptitude-cyclicity-remainders-a-8795869
the cyclicity of remainders when 128^n is divided by 9 is=>(2,4,8,77,5,1) i.e 6
1000/6 gives remainder 4
from the cyclicity list,the remainder is the no. 4 term i.e 7
easy na???

Is this solution Helpfull? Yes (9) | No   

umair  (10 years ago)

(128)^1000%9=(2^3)^333*2%9
(8)^333*2 %9
(-1)333*2 %9
-1*2%9
-2%9=-2.....==7

==

Is this solution Helpfull? Yes (6) | No (2)  

rahul soni  (10 years ago)

no one xplained clearly....

Is this solution Helpfull? Yes (5) | No   

DIVYA KUKRETI  (10 years ago)

ans 7
(9*14+2)^1000/9
=2^1000/9 (remainder theorem)
=(2^3)^333.2/9
=(9-1)^333.2/9
=-1.2/9
=-2/9
=7

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Guru Pranav  (10 years ago)

7

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

Bishnu Bhagat  (10 years ago)

128^1000=2^7000
Apply Fermit's Theorem i.e.
2^3=-1(mod9)
(2^3)^2333=(-1)^2333(mod9)
2^6999*2=-1*2(mod9)
2^7000=-2(mod9)
so the remainder is -2

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miraj ansari  (9 years ago)

128^1000 % 9

(126+2)^1000

2^1000 = (8^333)*2= (-1)^333*2 = -2%9=7

7 is the req. ans as simple as that....
:-}

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RAJA RAVI THEJA KALLA  (9 years ago)

128^1000 can also be written as 2^7000 and i am spitting 2^7000 as 2^6999*2 and 2^6999=8^2333 so 8^2333*2/9 is equal to -1^2333*2 so remainder is -2 and when prompted we get 9-2 7 as remainder

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vinita kumari  (10 years ago)

(64*2)^1000/9
remainder depends upon 2^1000/9 = (((2^6)^166)*2^4)/9
now the remainder depends upon (2^4)/9 = 7

Is this solution Helpfull? Yes | No (2)  

Ajay Poshak  (10 years ago)

Remainder 2
As 128 by dividing 9 leaves remainder 2 So
2^1000 divided by 9 leaves remainder 2.

Is this solution Helpfull? Yes | No (4)  

Guru Pranav  (10 years ago)

7 will be the answer !!!

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saurav roy  (10 years ago)

7(by remainder theorem)

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Guru Pranav  (10 years ago)

7 is the remainder

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Pratyusha Bhadra  (10 years ago)

7 is the answr....

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Manasa Nikitha  (10 years ago)

please xplain clearly

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saikat dhar  (10 years ago)

its 49
the 16 year is calculated on the four peaple.just because of every peaple cross 4 year according to thier ages,so avg age=(45*4+16+49)/5=49

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Suman Saurabh  (10 years ago)

(128)^1000/9=2^1000/9=(2^6)^166/9 * 2^4/9=2^4/9=7
so remainder=7

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techknowxpress  (10 years ago)

Ans=0
128^1000 stands for 128 XOR 1000= 720
720%9=0

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RITESH KUMAR  (9 years ago)

128%9=2
so we have to found 2^1000 %9
now 2^3 %9=-1
so 2^1000 %9 = (2^999)2%9 = ((2^3)^333)2%9=((-1)^333)2%9= (-2)%9=7
so ans=7

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kumari neetu sahu  (9 years ago)

any one please explain it clearly.........

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Deep  (9 years ago)

Abe Bah#n Ch%do agar ans short may likhna ha to eha pay maat likhoo...lerkibazi karne ka jaga nehii hain . focusii kar ra ha or idhar dimak garam ho raha hay.

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