Quant Boosters  Hemant Malhotra  Set 9

RHS is an integer and LHS terms are either positive or 0, that means all of the LHS terms need to be integers.
let a = x ^2 , b = y^2 and c = z^2
x + y + z = 25
[a+b+c+....r terms = n , whole number solutions of such equation is n+r1Cr1]ordered = 27C2 = 351
as for every x there will be a x^2 and that will be the value of a.Unordered: Values allowed are 025 for x,y,z
Out of these 351 ordered solutions, solutions are either
(1) x,y,z all distinct , so ordered had 3!, i.e. 6 solutions for one set of x,y,z
(2) x,y,z, two similar and third distinct, so ordered will have 3!/2!, i.e 3 solutions for one set of x,y,z
(3) x,y,z all same... only one solution per set
but here in our case x,y,z can not be all same.
Now, two same , & one distinct :
0,0,25
1,1,23
.........12,12,1
so 13 sets
so, all distinct sets = [total solution  3 * (two same & 1 distinct)]/6
= 351  39/6 = 312/6 = 52
in all solutions which are unordered = 52 + 13 = 65

Q21) If N is a natural no. less than 100, then for how many values of N are the nos. 6N + 1 and 15N + 2 relatively prime?

HCF(6N+1,15N+2))
HCF(12N + 2, 15N + 2)
HCF(12N + 2, 3N)
HCF(12N + 2, 2)
HCF(6n + 1, 1)
so all numbers will give hcf as 1 so 99

Q22) How many Ways 199 can be represented as sum of 3 distinct positive integers ?

a + b + c = 199
we need positive integers so
a = a'+1
b = b'+1
c = c'+1
so a' + b' + c' = 196
now total cases 196+31c31=198c2
now this will include cases when all r different let those cases are x
and way to arrange them is 6 so 6x
now case when 2 are same and one is different
2a' + b' = 196
b' = 196  2a'
so 99 cases
and way to arrange them is 3!/2!=3
so 99 * 3 =
now when all are same
so 3a' = 196
a' = 196/3 whihc is not possible so no case
now total ordered = 6x + 3 * 99
so x = 198c2  3 * 99/6

Q23) A=333....333 (51 digits) and B= 666....666 (51 digits). Find 52nd digit (counting from right) in the product A * B
a) 9
b) 7
c) 2
d) 1
e) none of the above

Check out the pattern,
3 * 6 = 18 so 2nd right digit is 1
33 * 66 = 2178 so 3rd right is 1
same in 51 digits, 52nd will be 1

Q24) What will be remainder when (67^67 + 67) is divided by 68

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