# Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 1

• OA : CBD
15 can be broken into three factors as follows - (5, 3, 1) or (15, 1, 1).
Solving for p, q, r we get (4, 3, 2) and (8, 8, 1) as the triplets corresponding to the above set of factors.
Thus product of p, q and r can be either 24 or 64.

• Q11) Sum of two numbers and their LCM is p where p is a prime and 10 < p < 30. For how many p do we have 3 pairs of such number?

• Let the numbers be ha and hb.
LCM = hab since LCM * HCF = ha * hb
So,
h(a + b + ab) = p
Case 1 : h != 1
Then, h = p => prime => Not possible
Since LCM >= Numbers >= HCF always
Case 2 : h = 1
a + b + ab = p
(a + 1)(b + 1) = (p + 1)
Number of pairs = [Factors of (p + 1)]/2 - 1 for the case when a becomes zero.
For the given range,
p = 11, 13, 17, 19, 23, 29
p = 11 => 12 = 2^2 * 3 => 3 * 2/2 - 1 = 2 pairs
p = 13 => 14 = 2 * 7 => 2 * 2/2 - 1 = 1 pair
p = 17 => 18 = 2 * 3^2 => 2 pairs
p = 19 => 20 = 2^2 * 5 => 2 pairs
p = 23 => 24 = 2^3 * 3 => 4 * 2/2 - 1 = 3 pairs
p = 29 => 30 = 2 * 3 * 5 => 2 * 2 * 2/2 - 1 = 3 pairs
So 2 such p = 23,29

• Q12) A natural number N is having a total of 21 composite factors. Let X be the maximum number of prime factors of N then find the remainder when X is divided by 3.

• X = 2. Hence remainder X mod 3 = 2

• Q13) Find the sum of all possible values of x such that 2/(sqrt(x)) + 1/(sqrt(y)) = 1/(sqrt(2)) where x and y are positive integers

• u = sqrt(x) and v = sqrt(y)
2v + u = uv/rt(2)
uv - 2rt2 v - rt2 u = 0
v(u - 2rt2) - rt2(u - 2rt2) = 4
(v - rt2)(u - 2rt2) = 4 = rt2 * 2rt2 = 2rt2 * rt2
u,v = 4rt2,2rt2 ; 3rt2,3rt2
x,y = 32,8 ; 18,18
Sum = 50

• Q14) How many three digit numbers are 34 times of sum of their sum of digits?

• 100a + 10b + c = 34a + 34b + 34c
66a - 24b - 33c = 0
66a = 24b + 33c
a = 1 ; b = 0 ; c = 2
a = 2 ; b = 0 ; c = 4
a = 3 ; b = 0 ; c = 6
a = 4 ; b = 0 ; c = 8
Four such numbers.

• Q15) Find the number of integers, N such that (N^2 + 2N - 8 )/(N^2 + N - 12) is an integer.

• N^2 + 2N - 8 = (N - 2)(N + 4)
N^2 + N - 12 = (N - 3)(N + 4)
=> (N - 2)/(N - 3) is an integer
=> (N - 3 + 1)/(N - 3) is an integer
=> 1 + 1/(N - 3) is an integer => N = 4. Only one possible value.

• Q16) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2, 5 or 6.

• I have two approaches for this one :
The Shareef approach :smile:
The Tareef approach :stuck_out_tongue_winking_eye:

The Shareef approach :
_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
342 numbers added till 999
1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
3 0 _ _ : 49 added
3 1 _ _ : 49 added
3 3 _ _ : 49 added
3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
Total 294 more. Remaining 21
3 9 0 _ : 7
3 9 1 _ : 7
3 9 3 _ : 7
3939

The Tareef approach :
0 -> 0
1 -> 1
2 -> 3
3 -> 4
4 -> 7
5 -> 8
6 -> 9
1000th term : 1000 in base 7 = 2626 3939

• Q17) How many positive divisors of 100^10 end in exactly two zeroes?

• 100^10 = 2^20 * 5^20
2^18 * 5^18
19 + 19 - 1 = 37

• Q18) How many pairs of non negative integers exist,such that the difference between their product and sum is 72?

• ab - a - b = 72
ab - a - b + 1 = 73
a(b - 1) - 1(b - 1) = 73
(a - 1)(b - 1) = 73 = 1 * 73 = -1 * -73
a,b = 2,74 ; 74,2 ; 0,-72 ; -72,0
Non negative integer pairs = 1 unordered or 2 ordered
Total integral solutions = 2 unordered or 4 ordered

• Q19) The equation 3^(x - 1) + 5^(x - 1) = 34 has :
a) No solution
b) 1 solution
c) 2 solutions
d) More than 2 solutions

• 34 = 25 + 9
x = 3 is the only solution
Both are exponentially increasing.
3 + 5 = 8< 34 ; 9 + 25 = 34 ; 27 + 125 > 34. No other values. Option b

• Q20) For what smallest positive integer N, will the number 3! + 5! + 7! + .. + (2N + 1)! be a perfect square ?

61

61

61

63

61

34

51