Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 5

Q27) Total number of real solution pair (x, y) if x = x^2 + y^2 and y = 2xy

x = x^2 + y^2
y = 2xy = 2y * (x^2 + y^2) = 2yx^2 + 2y^3
2y^3 + (2x^2  1)y = 0
y(2y^2 + 2x^2  1) = 0
y = 0 => x(1  x) = 0 => x = 0,1
OR
2(y^2 + x^2) = 1 => 2x = 1 => x = 1/2 => y = +1/2
4 solutions

Q28) Find the total number of digits in the sum of 1^1 + 2^2 + 3^3 + 4^4 + .... + 500^500
(log 2 base 10 = 0.30103)
a. 1348
b. 1349
c. 1350
d. 1500

The sum will have same no of digits as 500^500
[500 log 500] + 1 = 500(log 125 + log 4) + 1
= 500(3log5 + 2log2) + 1
= 500(3log10  3log2 + 2 log2) + 1
= 500(3  0.301) + 1
= (1500  150.xx) + 1 = 1350

Q29) Find the last two digits of 2^1  2^2 + 2^3  2^4 + 2^5 ... + 2^2013
a) 22
b) 42
c) 62
d) 82

2((2)^2013  1)/(3) = (2 + 2^2014)/3
(2^2014 + 2) = Q * 3 + 0
84 + 2 = Q * 3
Q => 62

Q30) N is a three digit number in the form of abc. If abc = a! + b! + c! then find the value of a + b + c ?

The numbers which are equal to sum of of factorial of its digits are called Factorion.
Only such known numbers till now are 1, 2, 145 and 40585.
So N = 145 and a + b + c = 1 + 4 + 5 = 10

If the same question is asked for say 3 or 5 numbers, how would the formula for pivot change? Could you please guide?

@sumit_agarwal I would urge you to consider a few examples yourself and try and work out first along the lines of above logic. Do it by standard approach first. It would help you develop a more generic intellect towards nonstandardised questions like these. Feel free to comment again if it doesn't work out. Will try and provide a solution for your query then.