# Quant with Kamal Lohia - Part 5

• Q1) Set S is formed by positive integers from 1 to 100 so that sum of no two elements of S is 103. What is the maximum possible number of subsets of S?

Basically we need to maximize the number of elements in S to maximize the number of subsets. So we can take exactly one element from each of brackets i.e. {1}, {2}, {3, 100}, {4, 99}, {5, 98}... {51, 52} i.e. 51 in all and thus number of subsets become 2^51.

Q2) In a community of 2029 inhabitants, at least how many have same initials for their first name and last name?

For example it is possible that at most all of 2029 inhabitants have same initials, AB (say)
Total different initials possible = 26 * 26 = 676. So considering WORST CASE, first layer of 676 inhabitants have all different initials, then second layer also, then third one. And finally one person remains that confirms that at least 4 persons are there which have same English initials for their first name and last name.

Q3) The number 3 can be written as 3, 2 + 1, 1 + 2, 1 + 1 + 1 in four ways. In how many ways the number "n" can be written?

Consider it as 1_1_1_1....n times. Now you just need to put + signs in place of _. There are total n-1 times _ is used and for each one you have two choices whether to put a + or not. That's why answer is 2^(n-1)

Q4) Find the unit digit of the expression 1^5 + 2^5 + 3^5 + … + 99^5

Unit digit of each number repeats after every fourth power i.e. unit digit of 29⁵ = unit digit of 29 = 9 and so on.
So unit digit of 1⁵ + 2⁵ + 3⁵ + ... + 99⁵ = unit digit of (1 + 2 + 3 + ... + 99) = unit digit of 99(99+1)/2 = unit digit of 99(50) = 0

Q5 ) a military general needs to take his troops of 100 soldiers across a river from bank A to bank B. He engages a boat with two boys, both of whom can row, at the bank A. But the boat can take only up to two boys, or only one soldier. What is the min. number of round trips that the boat has to make to transfer all the 100 soldiers and the general to bank B and come to bank A?
a) 400
b) 200
c) 202
d) 403

Here we are assuming that each soldier and the general too can row the boat. Initially all 100 soldiers, one general, 2 boys are at bank A. In first trip both boys go from A to B and one boy stays at B and other comes back to A in second trip. That completes 1 round trip. In next trip, one soldier moves from A to B and in the return journey, other boy returns from B to A and thus completes 2nd round trip. Thus in 2 round trips, one soldier is transferred from A to B and the boat is back at A. Thus it'll take exactly 2 * 101 = 202 round trips to take all 100 soldiers and a general to take from A to B and back the boat at A.

Q6) What will the place value of the first non-zero digit form the right hand side of "N" where N = ( 10^10) ! and " ! " is factorial?
Concept : Number of trailing zeroes in N! = number of powers of 5 in N! = [N/5] + [N/25] + [N/125] + [N/625] + [N/3125] +.... where [x] is greatest integer less than or equal to x, for example [5.1] = 5, [-0.1] = -1 and [4.9] = 4
[10¹⁰/5] + [10¹⁰/25] + [10¹⁰/125] + ... + 1 = 2000000000 + 400000000 + 80000000 + 16000000 + 3200000 + 640000 + 128000 + 25600 + 5120 + 1024 + 204 + 40 + 8 + 1 + 1 = 2499999998.

Q7) 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.

Q8) 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 ways

Now 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

Q9) How many pairs of 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 co-prime.
So required number of pairs are = {(2 * 2 + 1)(2 * 1 + 1)(2 * 1 + 1) - 1}/2 = 22.

Q 10) How many sets of three distinct factors of the number N = 2^6 × 3^4 × 5^2 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?

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

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