2014/031) A question on probability

I was walking through the park when I bumped into an old acquaintance.

He told me that he had two children. What is the probability that he has two girls?

A girl joined us. He said it was one of his children. What is the probability that he has two girls?

He said she was his eldest child. What is the probability that he has two girls?

Answer
probability that both are girls = 1/2 & 1/2 = 1/4

given that one is a girl it does not matter whether elder or younger probability that both are girls = probability that 2nd one girl = 1/2

2014/030)Show that the trigonometric equation sin(cos a)=cos(sin a) has no solutions.

Solution

if sin (x) = cos (y) then

x+y=π/2+2npi or (2n+1)pi−π/2

so lowest | cos a + sin a | = π/2

 but |cos a + sin a | <= (√2)

so no solution

2014/029) Given

2cos p+6cos q+7cos r+9cos s=0 and

2sin p−6sin q+7sin r−9sin s=0

Prove that 3cos(p+s)=7cos(q+r)

Solution

we have

2cos p+9cos s=−6cos q−7cos r

square and get


4cos^2 p+81cos^2 q + 36 cos p cos s=36cos^2 q +49cos^2 r −84cos q cos r...(1)

from 2nd relation given

2sin p−9sin s=6sin q−7sin r

square and get

4sin^2 p+81sin ^2 s - 36 sin p sin s=36sin ^2 q +49 sin ^2 r −84sin q sin r...(2)

and (1) and (2) to get
4(cos^2 p + sin^2 p) +81(cos^2 s + sin ^2 s) + 36 (cos p cos q- sin p sin s)=36(cos^2 q + sin ^2 q )+49 ( cos^2 r + sin ^2 r) + 84s( cos q cos r - sin q sin r)
or
4+81+36cos(p+s)=36+49+84cos(q+r)

or
36cos(p+s)=84cos(q+r)

or 3cos(p+s)=7cos(q+r)

2014/028) solve for real a (a+5)(a+4)(a+3)^2(a+2)(a+1) = 360

putting a + 3 =t we gen

(t+2)(t+1)t^2(t-1)(t-2) = 360

or rearranging the terns and multiplying we get

t^2(t^2-1)(t^2-4) = 360

putting t^2= x we get

x(x-1)(x – 4) = 360 (1)(

so x^3-5x^2 + 4x – 360 = 0

as 360 = 9 * 8 * 5 so from (1) x = 9 is a root

so we get factoing ( x – 9)(x^2 + 4x + 40) = 0





so x = 9 => a + 3 = 3 or – 3 so a = 0 or – 6

x^2 + 4x + 40 = 0 gives complex solutions

2014/027) Solve in natural numbers 2^x – 3^y = 7

x cannot be odd because if x is odd then

2^x mod 3 = -1 so 2^x – 3^y mod 3 = -1 so it cannot be 7 as 7 = 1 mod 3

y cannot be odd as if y is odd 3^y =3 mod 8

so 2^x – 3y = 5 mod 8 for x >= 3

if x = 1 or 2 2^ x < 7 so 2^x – 3^y = 7 not possible

so x and y both are even

say x = 2a and y = 2b

so 2^2a – 3^2b = 7

or (2^a + 3^b)(2^a- 3^b) = 7

so 2^a + 3^b = 7 and 2^a – 3^b = 1

solving these 2 we get a = 2 and b = 1 or x= 4 and y = 2

2014/026) How many ordered pairs of positive integers (x,y) satisfy:

x^2 + 10! = y^2

Solution

10! = 2 * 3 * 2^2 * 5 * (2 * 3) * 7 * (2^3) * 3^2 * ( 2* 5)
= 2^8 * 3 ^ 4 * 5^2 * 7

(x+y)(x-y) = 2^8 * 3 ^ 4 * 5^2 * 7

x and y both are integers when x + y and x -y both are even or odd

both cannot be odd

so both are even

no of factors of (x+y) *(x-y) = (8+1)(4+1)(2+1)(1+1) = 270

from this we need to remove the number of odd factors

(4+1)(2+1)(1+1) = 30
no solutions = 240

2014/025) If p, q and r are prime, and p divides qr − 1, q divides rp − 1, and r divides pq − 1, find all possible values of pqr

all 3 are different because if p = q then p divides pr-1 and

as p divides pr so p divides 1. so p is not prime. Without loss of generality we can assume p < q < r.

Pqr divided (pq-1)(qr-1)(rp-1)

so (pq-1)(qr-1)(rp-1)= mpqr where m is integer

so pqr – (pq^2r+p^2rq + pqr^2) + (pq + qr + rp) – 1 = mpqr

or pqr(1-p-q-r -m) = 1- (pq + qr + rp)

or pqrn = (pq + qr + rp) – 1 where n = p+q+r+m – 1

or dividing by pqr we get

1/ p + 1/q + 1/r = n + 1/(pqr)

as 1/p > 1/(pqr) we have n > 0

as 1/ p + 1/q + 1/r < 3 * (1/2) so n < 3/2

or n = 1

so 1/p + 1/q + 1/r = 1 + 1/(pqr) or > 1 ... (1)

p > 1 and p < 3 as 3 * 1/3 = 1 and so if p = 3 then 1/p + 1/ q + 1/r < 1

so p = 2 and q < 4 as 1/2 + 1/4 + 1/5 < 1

so q = 3

from (1) we get p = 2, q =3, r = 5 or pqr = 30

2014/024) Given a and b natural numbers

Given a and b natural numbers

(3a + b)^2 + 6a – 2b = 1544

find a +b



Solution

(3a + b)^2 +  6a – 4b = 1544

add 4b+1 to both sides to get

(3a + b)^2 + 2(3a + b) + 1 = 1544 + 4b

or (3a + b + 1)^2 = 1545 + 4b

as 1545 mod 4 = 1 solution may exist


so we need to take odd squares above 1545

(3a+b+1)=41=>1545+4b=1681=>b= 34

this gives a = 2 or a + b = 36

(3a+b+1)=43=>1545+4b=1849=>b=76 ( it should be be < 3a + b + 1 for being positive)

so a+b = 36 is the only solution

 

2014/023) solve the equation floor((25x-2)/4)= (13x+4)/3

(13x+4)/3 = integer say n

So x = (3n−4)/ 13
So we get
floor(25((3n−4)/13)-2)/4 = n

or 75 n – 126 > = 52n and < 52( n + 1)

or 23n > 126 and 23n < 178

or n = 6 or 7

or x = 14/13 or 17/13

2014/022) given

xyz = a/2..(1)

x^2+y^2 + z ^2 = a^2+ 6...(2)

x+y + z = a ...(3)
find 1/(xy + az) + 1/(yz + ax )+ 1/(zx + ay)

Solution

we have
xy + az = xy + z (x + y + z) = (z+x)(z+y) = (a-y)(a-z)
similarly
yz + ax = (a-y)(a-x)
zx + ay = (a-z)(a-x)

so 1/(xy + az) + 1/(yz + ax )+ 1/(zx + ay)
= 1/ (a-y)(a-z)+ 1/ (a-y)(a-x) + 1/ (a-z)(a-x)
= (3a – x – y – z)/)(a-y)(a-z)(a-x))


Numerator = 3a – a ( from (3)) = 2a

Denominator = a^3- (x+y+z)a^2 + (xy + yz + zx) a – xyz
= a^3- a. a^2 + (xy + yz + zx) a – xyz
= (xy + yz + zx) a – xyz

We have 2(xy + yz+ zx) = (x+y+z)^2 – (x^2 + y^2 + z^2) = a^2 – (a^2 + 6) = - 6

Or xy + yz + zx = - 3

So denominator = -3a – a/2 = - 7a/2

So value = - 4/7

2014/021)Solve for real solutions of the following system:

2z+z^2x=x
2y+y^2z=z
2x+x^2y=y

Solution

Z cannot be +/-1 because if z = +/- 1 then 2+x = x is a contradiction. Similarly x and y cannot be -1.

So we get

x = 2z/(1-z^2)

y = 2y/(1-y^2)

z = 2x/(1-x^2)


letting z = tan a we get x = tan 2a, y = tan 4a, z = tan 8a

or tan 8a = tan a

so 8a = npi + a

or 7a = npi

so solution set

z= tan (npi/7), x = tan (2npi/7), y = tan (4npi/7)

for n = 0 to 6