Quant Boosters  Hemant Malhotra  Set 9

67 mod 68 = 1
so (1)^67  1
so 2
so 68  2 = 66 remainder

Q25) I went to 'murugan idly shop' to get packed breakfast for my guests.There were 4 types of idli available and i wanted to take atleast 1 pack of each and atmost 5 packs of any.If I bought a total of 10 packs of idly,in how many ways could I have been billed?
a) 80
b) 84
c) 68
d) 56

a+b+c+d=10
1+a'+1+b'+1+c'+1+d'=10
so a'+b'+c'+d'=6
here 0 < =a' < =4
6+41c41=9c3
now a'=5+a''
5+a''+b'+c'+d'=6
so a''+b;+c'+d'=1
so 4c3
same for all 4 so 4 * 4c3=16
so 9c316
9 * 8 * 7/3 * 2  16
84  16 = 68

Q26) If a and b are natural numbers with no common prime factor and c is the greatest common divisor of (a+ b) and (a^2 +b^2) then how many values can c take?
a) 0
b) 1
c) 2
d) 3
e) Cannot be determined

gcd(a,b) = 1.
a^2+b^2= (a+b)^22ab
now c will divide either 2 or ab
Case 1:c divided ab
so c either divides a or b ,If c divides a it must divide a as well
so c will divide a and b both so c=1
Case 2: c divides 2.
c=1 or 2
so value of c could be 1 or 2

Q27) Digital sum of 1! + 2! + ... + 10 !

1!+2!+3! mod 9=0
4!+5!+6!
4!*(1+5+6)
4!*12 mod9=0
7!+8!=7!(1+8 ) so mod9=0
9!+10! multiple of 9=0
so digital sum=9

Q28) a + b + c + d = 21, number of solutions such that a, b, c and d are distinct natural numbers.

Method 1 :
a=x+1
b=x+y+2
c=x+y+z+3
d=x+y+z+w+4
so 4x + 3y + 2z + w = 11
(6+5+3+2) + (4+3+1) + (2+1) = 27
So, 24*27 = 648 solutions
This is just simple countingMethod 2 :
a+b+c+d=21
so a'+1+b'+1+c'+1+d'+1=21
so a'+b'+c'+d'=17
so total solution =17+41c41=20c3
now we want all distinct cases
so remove cases
case1 when three values are equal
so a=b=c
so 3a+d=20
so a wiill vary from 1 to 6 so 6 soulution
and total ways to arrange them 4!/3!=4
case2 when two are equal
so a=b
2a+c+d=20
so 45 cases but this will include cases when three are equal so remove that cases so 39 cases a
and way to arrange them 4!/2!*2!=12
so 20c3  4 * 6  12 * 39 = 648

Q29) Consider a pair (a,b) of natural numbers satisfying a + b^2 +c^3 = abc where c is the greatest common divisor of a and b .Then , how many such pairs are possible?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6

let a = Mc, b = Nc, so that M and N are coprime.
Then: Mc + N^2c^2 +c^3 = MNc^3.
M + N^2c + c^2 = MNc^2
So c must divide M. Put M = M'c, then M' + N^2 + c = M'Nc^2.
So M' = (N^2 + c)/(Nc^2  1).
So M'c^2 = N + (N+c^3)/(Nc^2  1).
So (N + c^3)/(Nc^2  1) is an integer.
If c = 1, (N+1)/(N1) can only be an integer for N = 2 or 3. so (a, b) = (5, 2) and (5, 3).let c > 1
bcz (N + c^3)/(Nc^2  1) is an integer and positive, we must have N + c^3 > = Ng^2  1, so N < = (c^3 + 1)/(c^2  1). If c = 2, then N < = 3. Then N = 1 gives the solution (a, b) = (4, 2), N = 2 gives (N + c^3)/(Nc^2  1) nonintegral and hence no solution,N = 3 gives the solution (a, b) = (4, 6). and for c > 2 there will be no solution so number of values = 4

Q30) The positive integer N has exactly six distinct integer divisors inclusing 1 and N. The product of five of these is 648. Which of the following must be a divisor of N
a. 4
b. 9
c. 12
d. 16
e. 24

Method 1 
Product of factors = N (number of factors/2))
let 6th factor = k
so k * 648 = N^3
so k * 3^4 * 2^3 = N^3
RHS is prefect cube so LHS shuld be perfect cube
so k = 3^2 = 9Method2 : N = 2 * 3^2
648 = 1 * 2 * 3 * 6 * 18
remaining 9 so OA = 9

@hemant_malhotra I don't understand how 248 came