# Quant Boosters by Hemant Malhotra - Set 19

• Find number of integral solutions of 2/x + 3/y = 1/10

Concept:
a/x + b/y=1/k
where a, b and k are positive integers and we want number of value of (x,y) satisfying this equation
Approach - first find number of factors of a * b * k^2 let number of factors=F
a) total number of positive integral solutions=F
b) total integral solutions = 2 * F-1
c)total number of negative solution= zero ( bcz if both x and y will be negative than lhs will be negative but rhs is positive so not possible )

solution:

First find total number of factors of 2 * 3 * 10^2
2^3 * 3 * 5^2 so (4 * 2 * 3) = 24 total number of factors
so positive integral solutions = 24
total number of integral solution = 2 * 24 - 1 = 47

Find the smallest positive integer such that (n-13)/(5n+6) is a non-zero reducible fraction.

hcf(n-13,5n+6)
hcf((n-13,5n+6-5n+65))
hcf(n-13,71))
so reducible form for min n=
n - 13 = 71
so n = 84

Find the minimum possible value of a + b + c if abc + bc + c = 2014, where a, b, c are positive integers.

c * (ab+b+1)=2014
c * (ab+b+1) = 2 * 19 * 53
so c=19
ab+b+1=106
ab+b=105
b * (a+1)=7*15
so b=7 and a=14
so sum =40

Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B, how many almond cookies are there ?

A gets twice So he has even number of cookies
even = even + even + even
even = odd + odd + even
three even is not here so he will choose two odd and one even
so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25
so 74 or 76 or 78 or 78 or 82
now B has 37 or 38 or 39 or 41
37 = 18+19 (19 is already used so not possible)
38 is not possible
39 = 18+21 (this is possible combination)
41 = 23 + 18
so 23 almond cookies are there

If x = 1! + 2! +3! +4! + ... + n!, how many value of n, x is a perfect square?

3,5,6 any perfect square will not leave this remainder when div by 7
so 1!+2!+3!+4!=33 mod 7=5
or 1!+2!+3!+4!+5!=153 mod 7=6
or 1!+2!+3!+4!+5!+6! mod 7=153+620 mod7=6+4=10 mod 7=3
so no value of n>=4 will give perfect square
so check for n=1,2,3
when n=1 then 1!=1 so perfect square
when n=2 then 1!+2!=3 not a perfect square
when n=3 then 1!+2!+3!=9=perfect square
so n=1 and 3 only two values

Find the number of integral solutions to |x| + |y| + |z| = 15
a. 902
b. 728
c. 734
d. 904

x + y + z = 15
so positive solutions = 15-1c3-1 = 14c2 = 7 * 13 = 91
now
when all negative =91
when 2 positive one negative 3c2 * 91
same for 2 negative one positive
91 + 91 + 3 * 91 + 3 * 91
8 * 91= 728
now case when one is zero
|y|+|z|=15
then
14 cases
again 14 * 4=56
so 3c2 * 56 = 3 * 56= 168
now when 2 are zero
|z|=15
so 2 solutions
so 3c2 * 2=6 solutions
so total 728+174 =902

If x² + xy +y² = 41, y² + yz + z² = 73 and z² + xz + x² = 9, find (x+y)/z.

(x^3 - y^3) = 41 * (x - y)
y^3 - z^3 = 73 * (y - z)
z^3 - x^3 = 9 * (z - x)
41x - 41y + 73y - 73z + 9z - 9x = 0
32x + 32y = 64z
so (x+y)/z = 2

Find all primes p, q, so that p^2 − 2q^2 = 1.

p^2-2q^2=1
(p^2-1) = 2 * q^2
(p-1)(p+1) = 2 * q^2
when p-1=1
then p+1 = 2q^2
so 2q^2=3 not possible
when p-1=2
then p+1=2q^2
4=2q^2 so q^2=2 no possible
when p+1=2 and p-1=q^2
p=1 and 0=q^2 not possible
now when p-1=2q
and p+1=q
so 2p=3q
so p=3q/2
so q=2 and p=3 will satisfy

Let F(x) =x^5+x^4+x^3+x^2+x+1 . Find remainder when F(x^12) is divided by F(x)

x^n-a^n=(x-a)(x^(n-1)+x^(n-2)+ .... a^(n-1)
f(x^12)=1+(x^12)+(x^12)^2+(x^12)^3+(x^12)^4+(x^12)^5
f(x^12)=1+(((x^12 -1)+1+(x^12)^2 -1 +1 +((x^12)^3 -1 +1 +(x^12)^4 -1 +1 +(x^12)^5 -1+1
f(x^12)=6+((x^12-1)+(x^12)^2-1+(x^12)^3 -1+(x^12)^4 -1 +(x^12)^5-1
now x^12-1=(x^6-1)(x^6+1)
x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)
so every term will contain x^6-1 and will be divisible by 1+x+x^2+x^3+x^4+x^5
so OA=6

The sum of first 2 terms of an infinite GP is 18. Also, each term in the series is seven times the sum of the terms that follow. Find the first term and common ratio of the GP

a = 7 * ( ar + ar^2 + ar^3 ... )
a=7(ar/1-r)
so a-ar=7ar
so 1-r=7r so 8r=1 so r=1/8
now a+ar=18
a(1+1/8)=18
so a * 9/8=18
so a=16

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.