Latest Elitmus Aptitude Question SOLUTION: find remainder when 33^34^35 divided by 7 ?

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find remainder when 33^34^35 divided by 7 ?

Asked In Elitmus himanshu uniyal (10 years ago)

Unsolved Read Solution (28)

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veer kumar  (10 years ago)

(33)^(34)^(35)/7
(28+5)^34^35/7 according to remainder theorm remainder will be 5^34^35/7
34^35 will be a even number
let write this in the form of 3
(34^35)/3=(33+1)^35/3 remainder will be 1.
so 34^35 can be written as 3k+1 (k can be anything doesnt matter)
as 34^35 is even so 3k+1=even=>k is odd
5^(3k+1)/7=(125^k)*5/7=(126-1)^k(0dd)*5/7
so remainder -1*5=-5
so make the remainder +ve 7-5=2answer

Is this solution Helpfull? Yes (37) | No (9)  

DIMPLE SINGH  (10 years ago)

this method is valid only for consecutive numbers..

rem((33^34^35)/7) = rem [{rem(33/7)+rem(34/7)+rem(35/7)}/7]
= rem [ (5+6+0)/7]
=rem [11/7]
=4
here rem== remainder

Is this solution Helpfull? Yes (18) | No (16)  

MOHIT JAIN  (10 years ago)

Remainder is 2

Is this solution Helpfull? Yes (12) | No (6)  

Shirsendu Bikash Das  (10 years ago)

33^34^35 is 33^1190. Now rem( 33^1190)/7= rem( 5^1190)/7=rem (-2^1190)/7
Now, -2^6 when divided by 7 leaves rem 1.
Therefore the said prob can be written as rem(-2^1188)/7*rem(-2^2)/7= 1*4=4
Rem= 4

Is this solution Helpfull? Yes (11) | No (13)  

Amit Kumar Pal  (10 years ago)

correct solution is 2
33^34^35/7=(-2)^4^5/7=(-2)^4/7=2

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ashish  (10 years ago)

ans is 2 if any doubt then check on == https://www.wolframalpha.com/input/?i=%2833%5E34%5E35%29mod7

Is this solution Helpfull? Yes (4) | No (2)  

Shashank  (10 years ago)

33^34^35 = 33^1190 = (28+5)^1190 = rem of (5)^1190 = rem of (25)^595 = rem of (21+4)^595 = rem of (4)^595= rem of (4^594) * 4 = rem of (64^198) * 4= rem of (63+1)^198 * 4 = rem of 1^198 * 4 = rem of 1*4 = 4

Is this solution Helpfull? Yes (3) | No (6)  

shyam kumar  (10 years ago)

(33^34^35)/7 can be written as (-2)^34^35/7 => 2^34^35/7
since 2^6 gives remainder of 1 when divided by 7.
try to write 34^35 in (6*k+r) form,
34^35, when divided by 6, gives remainder of 6.
so 34^35 can be written as (6*k+4),
thus, 2^34^35 can be written as 2^(6*k+4).
thus the remainder will be (2^4)%7=2.

hope this helps.

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

divyalakshmi  (10 years ago)

watch this remainder theorem concept in youtube for better understanding ...
http://www.youtube.com/watch?v=XEVuGq89tE0

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

Dilraj vyas  (10 years ago)

(33)^(34)^(35)/7

that means we can write (34)^(35)=N
so (35)^N/7=(5)^N/7
case 1:
5^1/7=5
5^2/7=4
5^3/7=-1
so we can get 1 at 5^6/7=1(remainder)
so cycle of 6 so N/6=34^35/6=(4)^35/6
we can also write as 4*4^34/6=2*(4)^34/3=2*(1)^34/3=2 or we have cancel 2 before so multiply
by 2=2*2=4
so the value of N is 4
and now (5)^4/7=-5 or 2 answer

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

Anuja Singh  (10 years ago)

remainder is 2

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

rakesh ranjan  (10 years ago)

(33^34^35)/7= (33^34^0)/7.... since 35/7=0, now 34^0=1
now, (33^1)/7= 33/7 so remainder will be 5

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

Ankit Giri  (10 years ago)

Ans is 0.

[(-2)*(-1)*0]/7
=> 0/7
=> 0.

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shyam kumar  (10 years ago)

sorry, in the above solution, 34^35 when divided by 6, gives remainder 4.

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kuldeep singh  (10 years ago)

remainder is 2
solve with eulers theorem

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Ehtesham  (9 years ago)

rem of (33^34^35)/7:

34 * 35=1190;


(33^1190)%7= (35-2)^1190%7
= (2^1190)%7
= (2^2*2^1188)%7
= (4*(2^3)^396)%7
= (4*(7+1)^296)%7
=4%7
=4

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

GAUTAM KUMAR  (9 years ago)

4
33^34^35=4

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sourav pal  (10 years ago)

33^34^35/7 =-2^-1^0(using negative remainder theorem)=2

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Kratica Rastogi  (10 years ago)

remainder is 2.

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sinny kedia  (10 years ago)

2 is d ans

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ASHISH KISHORE  (10 years ago)

(33)^(34)^(35)/7
now 34^35 divided by 3 remainder is 1
(98+1)^(34^35)/3
1^(34^35)/3
1^(3k+1)
1^3k+1
1+1=2 ans

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Rini  (10 years ago)

Remainder of (33,34,35) when divided by 7 are (-2,-1,0) .now multiply the remainders except 0 i.e. 2. So, 2 is the remaind.

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Nupur  (9 years ago)

The remainder is 2
35% 7 = 0
34% 7 = 6
33 % 6 = 5
5^6
30%7
2

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Himanshu Agarwal  (9 years ago)

unit digit of 33^34= 9
unit digit of 9^35=9
so,
9/7 gives the remainder as 2
so ans is 2

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Jayant Kumar Nayak  (9 years ago)

we can solve this easily using Euler's Theorem

x^e = x^(emod phi(n) )(mod n)

where ,
phi(n) = Euler Totient Function

if n can be prime factorised as follows:
n = p1^k1 * p2^k2 * ........ * pn^kn, where p1,p2,,,,pn are prime factors of n
then
phi(n) = n *(1-1/p1)*(1-1/p2)*.......*(1-1/pn)

if n is already a prime number then
phi(n) = n-1

we will use the above knowledge to solve 33^34^35 mod(7)

now we will go step wise

=33^34^35 mod(7)
so we get 33^k (mod 7) where k =34^35 mod(phi(7))
now k = 34^35 mod(6) as 7 is prme so phi(7) =6
k =4^35 mod(6)
which can be again broken down to k = 4^k1 , where k1 = 35 mod(phi(6))
so prime factors of 6 = 2 ,3
phi(6)=6 *(1-1/2)*(1-1/3)
= 2
therfore k1 =35mod2
k1 =1
k =4^k1mod(6)
= 4^1mod(6)
k=4
Now 33^kmod(7)
=(33mod(7))^4 mod(7)
=5^4 mod(7)
= 25mod(7) * 25mod(7) mod(7)
=4 * 4 mod(7)
=16mod(7)
=2

so remainder = 2

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kundan prasad  (9 years ago)

Ans : 2
sol : 33^34^35/7;
= (35-2)^34^35/7
=(-2)^34^35/7=(2)^34^35
=((2^3)^11^35*2^1^35)/7
= ((7+1)^11^35*2)/7 formula (a+1)^n=1
=2

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Pragati Priya  (9 years ago)

ans is 243

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Priyesh jain  (9 years ago)

33/7=5

5^34^35/7

cycle
5^1/7=5

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