# Quant Boosters - Sagar Gupta - Set 1

• Divisibility of 11
33 + a - (24 + b) = 0 or 11
9 + a - b = 0 or 11
9 + a - b = 11
a - b = 2 ......(1)
Divisibility of 9
57 + a + b = 6 or 15 .....(2)
a + b = 15 => a = 17/2 Not possible
a + b = 6 => a = 8/2 , b=4/2 => a,b=(4,2)

• Q15) Consider the increasing sequence 1, 3, 4, 9, 10, 12, 13… and so on. The sequence consists of all those positive integers which are powers of 3 or sum of distinct powers of 3. Find the 100th term of the sequence.

• Each number is of the form 3^0, 3^1, 3^1 + 3^0, 3^2, 3^2+3^0, 3^2 + 3^2 + 3^2+3^1+3^0
convert in base 3 :
1,10,11,100,101 and so on
100 in base 2 is 1100100
100th term will be 3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981

• Q16) There exist three positive integers P, Q and R such that P is not greater than Q, Q is not greater than R and the sum of P, Q and R is not more than 10. How many distinct sets of the values of P, Q and R are possible?

• q = p + x, r = q + y, where p is positive integer and x, y are non-negative integers
=> y + 2x + 3p < = 10
=> y + 2x + 3p' < = 7 (p' = p + 1)
So we have (4 + 3 + 1) + (4 + 2 + 1) + (3 + 2) + (3 + 1) + (2 + 1) + (2) + (1) + (1)
= 31 solutions

• Q17) India and Brazil play a football match in which India defeats Brazil 5-2. In how many different ways could the goals have been scored if Brazil never had a lead over India during the match ?

• First goal always India will score : 1-0
Now :
I / B B I I I I
Only 1 way brazil can take a lead :
( 6! / 2! 4! ) -1
14

• Q18) For any positive integer n, P(n) is the product of digits of n, then find the value of P(1) + P(2) + ...... + P(999).
Note:- P(1) = 1, P(6) = 6, P(23) = 6, P(900) = 9 and so on

• 0-9 : 1 + ( 1+2+3+.....9)
10-19 : 1 + ( 1+2+3+.....9)
20-29 : 2 + 2 ( 1+2+3......9)
.
.
90-99 : 9 + 9 ( 1+2+3...... 9 )
total =46 + 46 *45 = 46^2

100-109 : 1 + ( 1+2+3...9)
110 - 119 : 1 + ( 1+2+3+.....9)
120 - 129 : 2 + 2( 1+2+3......9)
.
.
190-199 : 9+ 9 ( 1+2 + 3 +.........9)
total = 46 + 46 *45 = 46^2

200-209 : 2 + 2 [ 1 + 2 +.......9 ]
210 -219 : 2 + 2 [ 1 + 2 +........9 ]
220 - 229 : 4 + 4 [ 1 + 2 +......... 9 ]
.
.
290 -299 : 18 + 18 [ 1 +2 +3 ...... 9
total = 92 + 92 [ 45 ] = 92 [ 46 ] = 2*46^2

If you see the pattern :
Series will go like :
46^2 + ( 46^2 + 2 * 46^2 +.........9 * 46^2 )
46^2 + 46^2 [ 45 ] = 46^3
Now subtract 1 from it as we considered 0 in the beginning
Answer = 46^3 - 1 = 97735

• Q19) The numbers a1, a2,...,a108 are written on a circle such that the sum of any 20 consecutive numbers equals 1000. If a1 = 1, a19 = 19, and a50 = 50, find a100

• HCF of 108 and 20 is 4, so terms will start repeating after every 4th term
So, a100 will be 1000/5 - 50 - 19 - 1 = 130

• Q20) Find the number of integral solutions of x^2 - 3y^2 = -2 , 0 < x < 20

• x^2 - 3y^2 = -2
x^2 = 3y^2 - 2
x^2 mod 3 = -2
x^2 = 3k - 2 = 3k + 1
3y^2 = 3k + 1 + 2 = 3(k+1)
y^2 = k+1
k = -1,0,3,8,15,24,35,48,63,80,99,120
x^2 = -2,1,10,25,46,73,106,145,190,241,298,361
x = 1,5,19 at k = 0,8,120 = y^2
Points : 1,1 ; 1,-1 ; 5,3 ; 5,-3 ; 19,11 ; 19,-11
6 solutions

• Q21) From an 'x' litre of solution of alcohol and water, 10% of the solution is removed and that amount of water is added back. Now, 9.09% of solution is removed and again water is added in the same amount. Again 8.33% of solution is removed and water is added in the same quantity. If initially, the mixture had alcohol and water in the ratio of 2:1, and has 1:1 alcohol and water afterwards, what can be the value of 'x' ?
a) 2 litre
b) 3 litre
c) 4 litre
d) Not dependent on x

• Suppose the original total is x, then alcohol as a fraction of total (step-by-step) becomes x * 2/3 * 9/10 * 10/11 * 11/12 = 1/2 and hence alcohol : water will become 1 : 1 always

• Q22) sqrt (x+2) + sqrt (x-2)= sqrt(4x-3)
The number of solutions for this equation are :
a) 0
b) 1
c) 2
d) more than 2

• x+2 + x-2 + 2 sqrt ( x^2 - 4 ) = 4x -3
2x - 3 =2 sqrt ( x^2 - 4)
4x^2 + 9 - 12x = 4x^2 - 16
25 = 12x
x=25/12

• Q23) N = 57^99 + 55^99. What is the remainder when N is divided by 224?

• Euler[224] =96
99 mod 96 = 3
57^3 + 55^3 mod 224
[57+55 ] ( 57^2 + 55^2 - 57*55 ) mod 224
112 * 3 mod 224 = 112

Or Use binomial
(56 + 1)^99 + (56 - 1)^99
Remainder will be 56 * 99 * 2 or 112(2 * 49 + 1) or 112

• Q24) Find the remainder when 987698769876.... 400 digits is divided by 31

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