ram wrote first 50 natural numbers on a black board.Then he erased two numbers sayp and q ,and replaced them by a single number N,he performed this operation repeatedlyuntil a single number was left.For all odd values of n,in nth operation,he choose N to bep+q+1,and for all even values of n he choose N to be p+q-1,find the final number whichremainedfind the remainder when 6^65^56 is divided by 43.
1st question is ambiguous....are we talking about the nth operation here?? pls clarify.
ram wrote first 50 natural numbers on a black board.Then he erased two numbers sayp and q ,and replaced them by a single number N,he performed this operation repeatedlyuntil a single number was left.For all odd values of n,in nth operation,he choose N to bep+q+1,and for all even values of n he choose N to be p+q-1,find the final number whichremainedfind the remainder when 6^65^56 is divided by 43.
@scrabbler if i take numbers to 1,2,3,4,5,6,7,8,9,10.....
then for odd (1 and 3) their summation is (3+1)+1 i.e 5 similarly for even ie (2+4)-1 i.e 5 so while traversing first time we may have +1 , -1 but in second traversing all values are coming odd 5,5,11(5+7+1),13(8+6-1),13 etc.
so m not getting how only +1 would be added finally ?
@scrabbler if i take numbers to 1,2,3,4,5,6,7,8,9,10.....then for odd (1 and 3) their summation is (3+1)+1 i.e 5similarly for even ie (2+4)-1 i.e 5 so while traversing first time we may have +1 , -1 but in second traversing all values are coming odd 5,5,11(5+7+1),13(8+6-1),13 etc.so m not getting how only +1 would be added finally ?
It is for the nth operation n is even or odd, not for the sum being even or odd at that stage. That's why I clarified. Go check the original post, it has been updated...
So 1st op mein 1 adds, 2nd mein subtracts and so on. At least, that's my interpretation. regards scrabbler
It is for the nth operation n is even or odd, not for the sum being even or odd at that stage. That's why I clarified. Go check the original post, it has been updated...So 1st op mein 1 adds, 2nd mein subtracts and so on. At least, that's my interpretation.regardsscrabbler
han yehi hoga...otherwise the question is kind of weird...that is what i was trying to clarify
net effect +1 hoga...hence the answer 1276, which u already explained..
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments?
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments? a. 210 b. 5040 c. 6 d. 10 e. 0
0 since no two smaller add up to more than a third larger one. regards scrabbler
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments? a. 210 b. 5040 c. 6 d. 10 e. 0
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments? a. 210 b. 5040 c. 6 d. 10 e. 0
In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?(A) 1(B) 2(C) 3(D) 4(E) 6
3. try to place the fourth vertex opposite each given points and form the figure.
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments? a. 210 b. 5040 c. 6 d. 10 e. 0
zero! fibonacci series... a+b=c but for triangles to exist.. a+b>c
Theory: If M and N are two Numbers co-prime to each other, i.e. HCF(M, N) = 1 and N=a^p*b^q*c^r...., Remainder[M^o(N)/N] = 1, where, o(N)= N(1 - 1/a)(1 - 1/b)(1 - 1/c)... and is known as Euler's Totient Function..o(N), or, E(N) is also the number of Numbers less than and prime to N..
Given, Rem[6^65^56/43] =?
Nw, E(43) = 42(1 - 1/43) = 42..
Thus, Rem[6^42/43] = 1..Since the power is 65^56, we will have to simplify this powers in terms of 42k + r. Therefore, we need to find the remainder when 65^56 is divided by 42..
Given 6 segments whose length are the elements of the set S={2,3,5,8,13,21} , what is the number of distinct triangles that can be formed using any three of these segments? a. 210 b. 5040 c. 6 d. 10 e. 0
How long is the side of the largest equilateral triangle that can be inscribed in a square whose side has length 1?A. 1B. √5/2C. 3√5/4D. 2-√3E. √(8-4√3)Please show your approach.
@scrabbler isnt option E less than 1.. its approx .8 in value..
Will edit...