Latest Elitmus Aptitude Question SOLUTION: Remainder when 33^34^35 divided by 11

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Remainder when 33^34^35 divided by 11

Asked In Elitmus shuvaranjan dash (10 years ago)

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RAKESH  (10 years ago)

33^34^35/11
=> (11*3+0)^34^35/11
=> rem(0^34^35) = 0
remainder = 0

Is this solution Helpfull? Yes (60) | No (3)  

SUDHIR SHARMA  (10 years ago)

Correct answer is 2
Because when when we solve 34^35 it will give 4 as a unit digit.
Therefore when 33^4 is divided by 7 it will give 2 as a remainder.

Is this solution Helpfull? Yes (3) | No (7)  

shilpi  (10 years ago)

zero , coz 33 itself is a factor of 11

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bithika das  (10 years ago)

a/c to remainders theorem 33^34^35=(0)(1)(2)=0
so ans is zero

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sajeesh c  (10 years ago)

i think the answer is zero.33^34^35 will be a multiple of 11

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Paras Sachdeva  (10 years ago)

0 because 33 divide by 11 gives 0 remainder

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PRATIM JANA  (8 years ago)

for any prime p,
(a^p) leaves a remainder of 'a' upon division by 'p'.

here p = 11
and a = 32

we have 32^(11*3)^34
and clearly 33^34 leaves a remainder of 0 when divides by 11.
thus we can write:
32^(11k) for some integer k.

using fermat's little theorem,
32 leaves a remainder of 10 upon division by 11
the answer is 10

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Surya Narayana K  (10 years ago)

Answer is 0.
Given that (33^34^35)/11
=>(33/11)^34^35
So Remainder is Zero.

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Vikash jain  (10 years ago)

Remainder is '0'
33^34^35=
first of all 35/34 remainder is 1 than
33^1=>33
al last 33/11=3 and remainder is zero(0).

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CHAITANYA  (10 years ago)

LET 34^35=X
33^X/11=(33^(X-1)*33)/11
11)33^(X-1)*33(3
33
-------------------
00----REMAINDER

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Mani kandan  (10 years ago)

HCF ( 33 , 11 ) is 0.
where phi(11)=(1-1/11)*11=10
=> 50^10k = 1mod11
So 34^35 is to be written in the form of 10k + a.
Now unit digit of 34^35 = 4 => 34^35 = 10k + 4.
=> 33^34^35 = 33(10k + 4) mod 11= 33^4 mod11 =0

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