Quant with Kamal Lohia  Part 9

Q1) x, y & z are positive real numbers
if xyz (x + y + z) = 1
Find the minimum value of (x + y) (y + z)when PRODUCT of some positive real numbers is CONSTANT then their SUM takes the LEAST value iff all the real numbers, under consideration, are EQUAL.
Here given that xyz(x + y + z) = 1 i.e. constant
i.e. [xz][y(x + y + z)] = 1 i.e. constant
i.e. [xz][xy + y² + yz] = 1 i.e. constant
So their sum i.e. [xz] + [xy + y² + yz] will be LEAST when both terms are equal i.e. 1 each.
So Minimum value of [xz] + [xy + y² + yz] = 1 + 1 = 2
i.e. Minimum value of [xz + xy + y² + yz] = 2
i.e. Minimum value of [x(y + z) + y(y + z)] = 2
i.e. Minimum value of (x + y)(y + z) = 2.Q2) Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
Just make the four cases i.e. both coins show
(i) 0 head... in C(3, 0) * C(4, 0) = 1 * 1 ways
(ii) 1 head... in C(3, 1) * C(4, 1) = 3 * 4 ways
(iii) 2 heads...in C(3, 2) * C(4, 2) = 3 * 6 ways
(iv) 3 heads...in C(3, 3) * C(4, 3) = 1 * 4 waysNow to get the required probability just multiply (1/8)(1/16) with the sum of above four ways to get the answer as = (1 * 1 + 3 * 4 + 3 * 6 + 1 * 4)(1/8)(1/16) = 35/128
Q3) How many divisors of 2100 are there whose HCF is 35?
2100 = 2² * 3 * 5² * 7
35 = 5 * 7
now we need to distribute 2² * 3 * 5 among two numbers in such a way that they are coprime.
So required number of pairs are = {(2 * 2 + 1) (2 * 1 + 1) (2 * 1 + 1)  1}/2 = 22Q4) The number of ways in which 185900 can be expressed as the product of a pair of co prime factors is
a) 8
b) 16
c) 4
d) 11If a number contains 'n' prime numbers, then number of ways to write the number as product of two coprime numbers is = 2^(n1) .. logic is very simple: every prime number with all its powers has to be put in one of the two numbers completely and that can be done in exactly 2 ways. So 2 ways for each prime's pack leads to 2^n and they are all ordered pairs counted exactly 2! = 2 times. That's why required unordered ways will be = (2^n)/2 = 2^(n1)
Q5) How many sets of three distinct factors of the number N = 26 × 34 × 52 can be made such that the factors in each set has a highest common factor of 1 with respect to every other factor in that set?
First thing is that N = (2^6) × (3^4) × (5^2)
Second is it is better here to make cases.
(p₁, p₂, p₃)  6 × 4 × 2 = 48
(p₁p₂, p₃, 1)  3(6 × 4 × 2) = 144
(p₁, p₂, 1)  (6 × 4) + (6 × 2) + (4 × 2) = 44
i.e. total 48 + 144 + 44 = 236Q6) How many perfect square numbers divide 24² + 32² + 75² completely without leaving any remainder?
4² + 32² + 75² = 8^2( 3^2 + 4^2) + 75^2
= 8^2 * 5^2 + 15^2 * 5^2
= ( 8^2 + 15^2) * 5^2
= 17^2 * 5^2
hence four squares 17 * 17, 5 * 5, 85 * 85, 1 * 1Q7) Given that 2^2010 is a 606digit number whose first digit is 1, how many elements of the set S = (2^0, 2^1 2^2... 2^2009} have a first digit of 4?
The smallest power of 2 with a given number of digits has a first digit of 1, and there are elements of S with n digits for each positive integer n < = 605, so there are 605 elements of S whose first digit is 1. Furthermore, if the first digit of 2^k is 1, then the first digit of 2^(k+1) is either 2 or 3, and the first digit of 2^(k+2) is either 4, 5, 6, or 7. Therefore there are 605 elements of S whose first digit is 2 or 3, 605 elements whose first digit is 4, 5, 6, or 7, and 2010 – 3 x 605 = 195 whose first digit is 8 or 9. Finally, note that the first digit of 2^k is 8 or 9 if and only if the first digit of 2^(k 1) is 4, so there are 195 elements of S whose first digit is 4.
Q8) Both roots of the quadratic equation x^2 — 63x + k O are prime numbers. Find the sum of all possible values of k.
Observe that sum of the two prime roots IS odd that means one root has to be 2.
Let a, b be the roots, then ab = 2 x 61 = k is the only possible value.
So requires sum is 122Q9) The number 2^29 is a 9digit number with distinct digits. Which digit is missing?
Just work modulo9. The missing digit is 4.
Q 10) How many integral triplets (x, y, z) satisfy the equation x^2 + y^2 + z^2 = 1855 ?
Note that 1855 = 7 mode 8 while all perfect squares are 0, 1 or 4 mode8. So it is impossible for 3 squares to sum up to 7 mod8. So no solutions are there.