Quant Boosters  Kamal Lohia  Set 3

If 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)

Q26) 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 = 236

Q27) 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 * 1

Q28) 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.

Q29) 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 122

Q30) 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.