Quant Boosters  Rajesh Balasubramanian  Set 2

24^{20} = (2^3 x 3)^20 = 2^{60} x 3^{20}
Number of factors of this number = 61 × 21 = 1281
24^{15} = 2^{45} x 3^{15}
Number of factors = 46 × 16 = 736
Factors of 24^{15} will be a subset of factors of 24^{20}.
So, the number of numbers that are factors of 24^{20} but not of 24^{15} is nothing but
61 × 21 – 46 × 16
= 1281 – 736
= 545.

Q26) How many natural numbers less than 10^4 exist that are perfect squares but not perfect cubes?

Number of perfect squares less than 10000 = 99.
10000 = 100^2; so till 99^2 will be less than 10000.
So, there are 99 perfect squares less than 10000?
From these some numbers that are also perfect cubes have to be eliminated.
So, we are looking for numbers that are perfect squares and perfect cubes. Or, we are looking for powers of 6.
1^6, 2^6, 3^6, 4^6 are all less than 10000, but 5^6 is greater than 10000.
So, there are only 4 powers of 6.
So, out of 99, we need to subtract 4 possibilities. Or, there are 95 different natural numbers that will be perfect squares but not perfect cubes

Q27) Sum of three Natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?
a) 45
b) 36
c) 54
d) 28

a + b + c = 10. Now, let us place ten sticks in a row
         
This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance
    +      + 
This would be the equivalent of 4 + 5 + 1. or, a = 4, b = 5, c = 1.
There are 9 slots between the sticks, out of which one has to select 2 for placing the '+'s.
The number of ways of doing this would be 9C2. Bear in mind that this kind of calculation counts ordered triplets. (4, 5, 1) and (1, 4, 5) will both be counted as distinct possibilities.
We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural numbers as in this question) with a small change to the above approach. Give it some thought.

Q28) Sum of three Whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?
a) 66
b) 78
c) 72
d) 56

a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying. We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.
If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.
a + b + c = 13. Now, let us place ten sticks in a row
            
This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance
    +      +    ,
This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.
There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s.
The number of ways of doing this is 12C2.

Q29) In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?
a) 72
b) 54
c) 45
d) 36

This is nothing but the number of ways of having a, b, c such that a + b + c = 11, where a, b, c are natural numbers. By having them to be natural numbers, we ensure that no box can be empty (no zeroes)
The question is similar to the previous one.
Correct Answer : 10C2 = 45

Q30) a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?
a) 8
b) 6
c) 2
d) 4

Exactly one of ab, bc and ca is odd => Two are odd and one is even
abc is a multiple of 4 => the even number is a multiple of 4
The arithmetic mean of a and b is an integer => a and b are odd
and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3
c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9
Four triplets are possible.