# Quant Boosters - Quant Mixed Bag - Set 1

• [Credits : Sabyasachi Mishra]

Let’s say x= a^2 - b^2 = > x = (a+b)(a-b)

If x has to be written as the difference of squares of two natural numbers in only one way, then there must be only one way to write x as (a+b) * (a-b) where (a+b) > (a-b).
means that b can’t be zero and hence, (a+b) != (a-b)

Now, (a+b) and (a-b) will either both be odd or both be even.

1. Case : Both (a+b) and (a-b) are odd
x = odd * odd = > odd
If x is an odd number, it can always be represented as 1 * x. If x has more than one factor(excluding one and x itself), say two factors then x= c * d can always be written. So, we have to find x where x is either an odd prime or x is the square of an odd prime.
x= k^2 (k being an odd prime) can be factorized as 1* k^2 or k * k.
k*k would mean x = k^2 - 0^2 (not allowed as 0 is not a natural number)
We have 25 prime numbers less than 100, hence 24 cases(excluding 2) and 9,25,49 as squares of odd primes = > 27 cases in all
2. Case : Both (a+b) and (a-b) are even
x = even * even = > even
x = 2a * 2b = > x = 4 * (ab) & x < = 100
= > ab < = 25 where a!=b & a and b are both natural numbers
Lets say K=ab, K should be represented as ab in only one way. Similar to the above logic (except that numbers can be even too). K can be prime numbers or prime squares. So, k = 2,3,5,7,11,13,17,19,23 or 4,9
= > 11 cases in all

So there are 27 + 11 = 38 numbers between 1 and 100 which can be represented as the difference of natural squares in only one way.

• Q17) How many three digit numbers are there whose sum of digits is divisible by 5?

• [Credits : Sabyasachi Mishra]

Lets say the 3 digit no. is abc.

1. Case 1 - two of them are zero = > 1
Only one possible number 500.
2. Case 2- one of them is zero = > 4 * 8 + 2 * 1 =34
a can’t be zero. So, a + b or a+c is divisible by 5.
For a = 1, the other one could be 4 or 9
a = 2, the other could be 3 or 8
Except for a=5 rest each case has two possiblities = > 4 numbers each
For 5 ther’s 550 or 505 = > 2 numbers
3. Case 3- none of them is zero = > 6 + 36 + 61 + 36 + 6 = 145
Sum of all three digits can vary between 3 and 27 as none of them is zero.
So, the sum can be 5, 10, 15, 20 or 25
1) Sum is 5 = > 6 cases
1 + 1 +3 = 5 = > 3 numbers
1 + 2 + 2 = 5 = > 3 numbers
2) Sum is 10 = > 3 * 4 + 6 * 4 = 36 cases
1 + 1 + 8 = > 3 numbers
1 + 2 + 7 = > 6 numbers
1 + 3 + 6 = > 6 numbers
1 + 4 + 5 = > 6 numbers
2 + 2 + 6 = > 3 numbers
2 + 3 + 5 = > 6 numbers
2 + 4 + 4 = > 3 numbers
3 + 3 + 4 = > 3 numbers
3) Sum is 15 = > 15 + 18 + 18 + 9 + 1 = 61 cases
1 + (5 + 9) / (6 + 8 ) / (7 + 7) = > 6, 6, 3 numbers = 15
2 + (4 + 9) / (5 + 8 ) / (6 + 7) = > 6, 6, 6 numbers = 18
3 + (3 + 9) / (4 + 8 ) / (5 + 7) / (6 + 6) = > 3, 6, 6, 3 numbers = 18
4 + (4 + 7) / (5 + 6) = > 3, 6 numbers = 9
5 + 5 + 5 = > 1 number = 1
4. Sum is 20 = > 3 + 6 + 9 + 12 + 6 = 36 cases
2 + 9 + 9 = > 3 numbers = 3
3 + 8 + 9 = > 6 numbers = 6
4 + (7 + 9) / (8 + 8 ) = > 6,3 numbers = 9
5 + (6 + 9) / (7 + 8 ) = > 6, 6 numbers = 12
6 + (6 + 8 ) / (7 + 7) = > 3, 3 numbers = 6
5. Sum is 25 = > 3 + 3 = 6 cases
9 + 8 + 8 = > 3 numbers = 3
9 + 9 + 7 = > 3 numbers = 3

145 + 34 + 1 = 180
So, there are 180 such numbers.

There is also this easier alternate method that is a bit difficult to visualize if you are encountering such a question for the first time. But, here it goes. You can say that in every consecutive 10 numbers where only the ones place is changing, there would be exactly 2 such numbers whose digits’ sum is divisible by 5. I can have (100 - 109), (110 - 119) … (990 - 999) i.e. 90 such cases as there are 900 3 digit numbers. And hence, there would be 90 * 2 = 180 such numbers.

• Q18) How many natural numbers between 1& 100 have exactly 4 factors?

• [Credits : Sabyasachi Mishra]

Every number x can be represented in this manner.

x = (a1 ^ b1) * (a2 ^b2) * (a3^b3) … where a1, a2, a3 are all prime numbers

Number of factors of x = (1+b1) * (1+b2) * (1+b3) …

Now, we know that 4 = 2 * 2 or 1 * 4

So, all numbers which have 4 factors would either be in the form of

1. a^3 where a is a prime; Factors would be 1,a,a^2 & a^3
2. a * b where a and b are both primes; Factors would be 1,a,b & ab

Prime cubes between 1 and 100 are 8 and 27. So, 2 cases.

Cases

1. For a=2, b = 3,5,7,11,13,17,19,23,29,31,37,41,43,47 = > 14 cases
2. For a=3, b=5,7,11,13,17,19,23,29,31 = > 9 cases
3. For a=5, b=7,11,13 = > 3 cases

So total possible cases = 2 + 14 + 9 + 3 = 28

• Q19) What are the last two digits of 168^1057?

• [Credits : Sabyasachi Mishra]

Think of this as (160+8 )^1057.

When you expand this term, the last two digits of all terms would be 00 except 8^1057 and 1057 * 8^1056 *160. (Binomial expansion used)

The second term clearly would be …60, 0 due to 160 and 6 due to multiplication of 8^1056 = (8^4)^264 = 6^254 by 6.

For, the first term

8^1 = 08 Case 1
8^2 = 64
8^3 = 12
8^4 = 96
8^5 = 68 Case 2
8^6 = 44
8^7 = 52
8^8 = 16
8^9 = 28 Case 3
8^10 = 24
8^11 = 92
8^12 = 36
8^13 = 88 Case 4

So, we can say multiplying 8 by 8^4 adds 6 to tens place.

8^1057 = 8 * (8^4)^264 = > adding 6 264 times to the tens place i.e. 0 and keeping the units place as 8 = > last 2 digits are 48.

So, last 2 digits overall = 48 + 60 = 08

P.S. You can figure out the trend of last 2 digits by solving upto Case 3.
Also generally (x^4)^y follows a pattern for all x, where y keeps increasing by 1. So, you could directly see the trend for (8^4), (8^4)^2, (8^4)^3 and deduce from that.

• Q20) With a given amount X, Mr. Y can buy 20 E or 12 S or 5 P. If he purchased an equal number of E and S and P with an amount Z and given that Z = 3X, Find the number of Ps bought by him ?

• [Credits : Tanveer Iqbal]

Let X =100
Price of E/S/P - 5/8.33/20
Price of E+S+P= 33.33
Number of sets for X(100 ) = 100/33.33 =3
Number of sets fo 3X (300 ) = 3X3 = 9

Other Methods:

#1
Price ratio of P:S:E=12:5:3.
Consider X=60 = > Z=180 = > Total share of P=108. Number of P=108/12=9

#2
One can think of an analogous problem in Time work Domain:
P, S, E can finish a job in 5, 12, 20 days respectively. They are assigned to complete a work together which thrice as large as the previous one. In how many days the work will get completed if each one works with the same efficiency.

#3
Treat the first set of equations as amount of work done and time taken by each person respectively to complete the task
In the next set of equations, they all work together to complete a task working equal no.of days
We have to find that "equal" no .of days

• Q21) There are z digits in the decimal expression of the natural number N,while there are y digits in the decimal expression of N^3. Then which of the following cannot be equal to y + z?
(a) 20
(b) 26
(c) 35
(d) 45
(e) none of these

• [Credits : Tanveer Iqbal]
As per the question,
10^(z−1) < = N < 10^z = > 10^(3z−3) < = N^3 < 10^3z.
So, y can be {3z − 2, 3z − 1, 3z} and y+z can be {4z-2 ,4z-1,4z)
Clearly all the natural numbers except those of type 4k+1 satisfy the value for y+z.
So, 35 cannot be the value.
= > Choice (d) is the right answer

• Q22) A real number x is chosen at random in the interval [-21/2, 21/2]. What is the probability that the closest integer to x is odd?
a) 1/21
b) 5/11
c) 10/21
d) 11/21
e) 1/2

• [Credits : Tanveer Iqbal]
From [-10.5,10.5] There can be 21 intervals [-10.5,-9.5],[-9.5,-8.5].....[9.5,10.5]..and there are 10 intervals [-9.5,-8.5]....[8.5,9.5]..in which any x selected in those intervals is closest to an odd integer.
So the probability is 10/21
Choice (c) is the right answer

• Q23) A positive integer p is called "CAT number" if p^3 + 7p = q^3 + 133 for some positive integer q. The sum of all such "CAT numbers" is
(a) 24
(b) 26
(c) 30
(d) 34
(e) 38

• [Credits : Taveer Iqbal]
We consider the size of p relative to q. If p = q, then q is a CAT number if and only if
7q = 133. In this case, we get 19 is a good number. Now, suppose that p > = q + 1. Then
q^3 + 7q − 133 = p^3 > = (q + 1)^3 = q^3 + 3q^2 + 3q + 1
so that 3q^2 − 4q + 134 < = 0. For positive integers q, it is easy to see that this is impossible. It remains to consider p < = q − 1. For such p, we have q^3 + 7q − 133 = p^3 < = (q − 1)^3 = q^3 − 3q^2 + 3q − 1 which implies 3q^2 + 4q − 132 < 0. It follows here that q < 6. One checks that 6^3 + 7 · 6 − 133 = 5^3 and 5^3 + 7 · 5 − 133 = 3^3,
so 6 and 5 are CAT numbers. For q < = 4, we see that q^3 +7q−133 < 0 and, hence, q^3 +7q−133 cannot equal p^3 for a positive integer p. Therefore, the sum of all CAT numbers is 19 + 6 + 5 = 30.

• Q24) If P : Q = 1 / 2 : 3 /8 , Q : R = 1 / 3 : 5 / 9 and R : S = 5 / 6 : 3 / 4 , Then P : Q : R : S = ?

• [Credits : Tanveer Iqbal]

P : Q = 1 / 2 : 3 / 8 = 4 : 3 , Q : R = 1 / 3 : 5 / 9 = 3 : 5 and R : S = 5 / 6 : 3 / 4 = 10 : 9
P : Q = 4 : 3 , Q : R = 3 : 5 , R : S = 10 / 2 : 9 / 2 = 5 : 9 / 2
P : Q : R : S = 4 : 3 : 5 : 9 / 2 = 8 : 6 : 10 : 9

• Q25) What digit does “a” represent, if 35! = 10333147966386144929a66651337523200000000?
(a) 4
(b) 6
(c) 2
(d) 1

• [Credits : Tanveer Iqbal]
35!= 1 * 2 * 3 * 4 * 5 *... * 35
so the above number will also be divisible by 3,9,11( such number which uses the sum of number to check the divisibility)
Check by 3:- you will get the different values of a which also have to satisfy for 9,11.
Check by 9:- (Sum of digits +a) has to be divisible by 9.
Check by 11 directly,
Sum of odd places- sum of even places= 11 n or 0.
(66 + a) - 72=0, as a cant be greater than 9. to give the difference multiple of 11.
so a = 6

• Q26) In base 7, 65324 + 50231 - 133 × P = 130462. What is the value of 'P'?
a) 110
b) 101
c) 1001
d) 111

64

61

51

61

42

61

62