@abhishekj1991 said:find a remainder of 2^133 divided by 133?
128?
@shattereddream said:We will put 4 here and not 28 i think its clear now
we have
-> If f(x)*f(1/x) = f(x) + f(1/x) --------------- eq 1
-> If x = 3
-> substituting x = 3 in eq 1 since f(x) = X^n + 1
-> LHS => f(3)*f(1/3) => (3^3 + 1)*(1/3^3 + 1)=> 28*27/28
-> RHS => f(3) + f(1/3) => (3^3 + 1) + (1/3^3 + 1)=>28 + 27/28
LHS != RHS
Correct me If I am wrong
-> If f(x)*f(1/x) = f(x) + f(1/x) --------------- eq 1
-> If x = 3
-> substituting x = 3 in eq 1 since f(x) = X^n + 1
-> LHS => f(3)*f(1/3) => (3^3 + 1)*(1/3^3 + 1)=> 28*27/28
-> RHS => f(3) + f(1/3) => (3^3 + 1) + (1/3^3 + 1)=>28 + 27/28
LHS != RHS
Correct me If I am wrong
@abhishekj1991 said:find a remainder of 2^133 divided by 133?
128
E(133)equals to 108
so 2^108*2^25 /133
(2^7 )^3*2^4/133
(-5)^3 *16/133
deducing 128
E(133)equals to 108
so 2^108*2^25 /133
(2^7 )^3*2^4/133
(-5)^3 *16/133
deducing 128
say a has x kgs, b has y kgs. removing p kgs from A, mixing remaining portion of A,B. we have rice = 5(x-p)/8 +7(y)/10 =8,
dhal= 3(x-p)/8 + 3(y)/10=3
but if you put x-p=a in both equations and solve, you get negative values of y, a.
there seems to be an error in the question 
@IIM-A2013 said:Two Boxes A and B were filled with a mixture of rice and Dal - in A ratio of 5:3 and in B in the ratio of 7:3.What quantity must be taken from the first to form a mixture that shall contain 8kg of rice and 3 kg of dal ?
@IIM-A2013 said:Two Boxes A and B were filled with a mixture of rice and Dal - in A ratio of 5:3 and in B in the ratio of 7:3.What quantity must be taken from the first to form a mixture that shall contain 8kg of rice and 3 kg of dal ?
this is not possible
@IIM-A2013 said:Two Boxes A and B were filled with a mixture of rice and Dal - in A ratio of 5:3 and in B in the ratio of 7:3.What quantity must be taken from the first to form a mixture that shall contain 8kg of rice and 3 kg of dal ?
alligation principle fails 😛 😛 :P
@abhishekj1991 said:find a remainder of 2^133 divided by 133?
133 is a prime no
and according to fermat theorem
a^p /p where P is a prime no remainder will be a
so the answer is 2
and according to fermat theorem
a^p /p where P is a prime no remainder will be a
so the answer is 2
@ajeetaryans said:133 is a prime no
and according to fermat theorem
a^p /p where P is a prime no remainder will be a
so the answer is 2
133 = 7*19. So, it's not a prime.!
@ajeetaryans said:133 is a prime noand according to fermat theorema^p /p where P is a prime no remainder will be aso the answer is 2
133 is not a prime number
133 equals to 19*7
:)
133 equals to 19*7
:)
@ajeetaryans said:133 is a prime noand according to fermat theorema^p /p where P is a prime no remainder will be aso the answer is 2
bhai 133 is not a prime ..its div by 7.. you can say 2 & 133 are coprime and hence eulers can be applied
:)
:)
guys options are
4kg
5kg
6kg
cannot be achieved