Latest Elitmus Aptitude Question SOLUTION: What's the remainder if 32^32^32 is devided by 9 ? A)4 B)7 C)1 D)2

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(#M40018635) ELITMUS question Keep an EYE

What's the remainder if 32^32^32 is devided by 9 ?
A)4
B)7
C)1
D)2

Asked In Elitmus Prakash Gupta (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

nishanth  (10 years ago)

32^32^32=2^5120
use pattern method
1) 2^1/9=2
2) 2^2/9=4
3) 2^3/9=8
4) 2^4/9=7
5) 2^5/9=5
6) 2^6/9=1
after 6 steps pattern is repeating
7) 2^7/9=2

divide 5120 and see whats the remainder rem=8 after 6 steps 2 steps gives 4
so ans is a)4

Is this solution Helpfull? Yes (39) | No (4)  

sarab  (10 years ago)

can anyone explain how 32^32^32= 2^5120... ??

Is this solution Helpfull? Yes (12) | No (2)  

navneet kumar verma  (10 years ago)

32^32^32=2^5120
>> 2^2 * [2]^5118
>> 4* [2^3]^1706
>> 4* [8]^1706
>> 4* [9-1]^1706 using ((a-1)^n)/a= 1 if n is even, else a-1.
>> 4* [-1]^1706, because power is even, evaluation of brackets will give 1.
thus 4*1=4 Ans

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

yashpal  (10 years ago)

since unit place 2^2 = 4 so 2^32 = 4
||ey 2^4 = 16/9 = 7 remainder

Is this solution Helpfull? Yes (5) | No (3)  

mithlesh dhore  (10 years ago)

ans is D)2

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mithlesh dhore  (10 years ago)

u r right nishant the ans is 4 my mistake

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Anshul Kumar  (10 years ago)

see the unit place..
2^2^2= 16
16/9 = 7
Ans=>7

Is this solution Helpfull? Yes (1) | No (4)  

SAMBIT KUMAR GHOSH  (10 years ago)

what is the ans??pls give someone the right xplanation...my ans is 7

Is this solution Helpfull? Yes (1) | No (4)  

lucky  (10 years ago)

32^1=32
32^2=1024
32^3=----68
32^4=-------76
32^5=------32
so after that it goes on repeating like this.....
thus 32/5 gives two extra.....
hence last two digits of 32^32=68......
now 32^32^32=32^68 considering only last two digits....
now 68/5 gives 3 extra...
thus it will give 76 at last two places..
now 76/9 gives remainder..4

ans:- 4

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naga sarath chand  (10 years ago)

question can be written as
32^(32*32*.....32times)/9= 2^(5*32*32....32times)/9 =
[2^(32*32....32times)]*[2^(32*32......32times)]*....5times/9 (because 5 is there i wrote it 5 times seperately)
but 2^32*32*.....32times = 2^4n=2;
therefore
above big expression = 2*2*2*2*2/9 = 2^5/9 = 2*2^4/9 = 2*2/9 = 4/9 =4
(4 is the answer).

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Krishanu chatterjee  (10 years ago)

right ans will be 4

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naga sarath chand  (10 years ago)

sorry,
2*2^4/9 = 2^5/9 = 2/9.
so, the answer is 2.
sorry i have given it as in the above solution of myne as 4.please correct it.

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ratneshh  (10 years ago)

32^32^32=
((2^5)^32)^32=
(2^160)^32=
2^160*32=
2^5120
sarab this is the step

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Tapabrata Pal  (6 years ago)

32^32^32/9
=32^64/9
=2^5^64/9
=2^69/9
now 2^9=256
256/9 then remainder=4
answer=4

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swati  (6 years ago)

HCF(9,32)=1 so,both are co-prime
rem[32/9]=5
so rem[5^f(9)/9]=1
so f(9)=9*(1-1/3)=>6
rem[5^6/9]=1
now , rem[32^32/6]=4
so , 32^32=6n+4
so,rem[ 5^(6n+4)/9]=>rem[5^6n/9] *rem[5^4/9]
=>1 *rem[4/9]=4(ans)

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