# 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 co-prime numbers is = 2^(n-1) .. 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^(n-1)

• 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 606-digit 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.

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