Q13/110) Solve in positive integers (1+1/x) (1+ 1/y))(1 +1/z) = 2

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3

(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2

(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Q13/109) Find all positive integer solutions of the equation 4x^3 – 12x^2 + 5x – 10y + 36y^2 – 18y^3 + 4x^2y + 6xy – 15xy^2 = 0

We put the number in decreasing order of power of x

4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−10y+36y^2−18y^3
Now factor the part independent of x
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(5-18y+9y^2)
= 4x^3−12x^2 +4x^2y +5x+6xy−15xy^2−2y(3y-5)(3y-1)
Now I multiply by 2 and put 2x = z to get coefficient of x^3 as 1
= ½(8x^3−24x^2 +8x^2y +10x+12xy−30xy^2−4y(3y-5)(3y-1)
= 1/2(z^3−6z^2 +2 z^2y +5z+6zy−15zy^2−4y(3y-5)(3y-1)
= 1/2(z^3−z^2(6-2y) +z(5+6y+15y^2) −4y(3y-5)(3y-1)

Now we need to split - 4y(3y-5)(3y-1)into3 parts that the sum is 6- 2y

They are-( 3y-5),- (3y-1) , 4y ( it is easy to do so as 5 + 1 = 6

Now to check The coefficient of z is (5+6y+15y^2)

We see that – 4y(3y-5) – 4y(3y-1) + (3y-1) (3y-5) = -15y^2 + 6y + 5 which is true

So we get ½(( z- 4y) ( z+ 3y-5)(z+ 3y-1)) or (x-2y)(2x + 3y-5)(2x+3y-1)) = 0

Which gives us x = 2y or 2x + 3y = 5 or 2x + 3y = 1

X= 2y gives solution x= 2t, y = t;

2x + 3y = 5 gives x= 1, y= 1 and 2x + 3y = 1 has no solution ( as we are looking positive solution)

Q13/108) Evaluate the given sum k from 0 to n ∑ (nCk) /(k+1)

(nCk)/(k+1) = n!/(k! * (n-k)!) * (k+1) = 1/(n+1) ( n+ 1 C k+1)

So ∑ (nCk)/(k+1) = 1/(n+1) (∑ (n+1 C k) - (n+1 C 0))
= 1/(n+1) (∑(n+1Ck) - 1)
= 1/(n+1) ( 2^n+1) – 1)

Q13/107) What is the general formula for the equation 1*1!+2*2!+.....+n*n!

We have k* k! = ((k+1)-1) * k! = (k+1)! – k!

So 1 * 1! = 2! – 1!

2 * 2 ! = 3! – 2!
n*n! = (n+1)! – n!

Adding we get 1*1!+2*2!+.....+n*n! = (n+1)! – 1! or (n+1)!- 1

Q13/106) Find three different prime numbers between 10 and 99, where the average of any two is a prime and the average of all three is a prime.

All the numbers have to be of the same form that is all 6n+1 or all 6n-1.
This is so because 6n+1 and 6m-1 have average 3(n+m) which is not a prime
this I took to reduce the number of test cases.
Now without loss of generality we can choose the 2 numbers and their mean to be taking
a, a + 6n, a + 12n, all 3 are primes ( a and a + 12n are numbers and a+6n is mean we choose this because average should be 6 or a multiple)
the next number has to be of the form a + 12m 
now (a+12n + a + 12m)/2 = a + 6(m+n)
and (a+ a + 12n + a + 12m)/3 = a + 4(m+n)
so the 7 prime numbers are a, a + 6n, a+ 12n, a+12m, a+ 6(n+m), a +6m, a + 4(m+n)
not necessarily in increasing order

a=11, n= 3, m= 5 gives the solution 11,29,47,71, 59, 41, 43.

Q13/105)if ab + bc +ca=0,then find 1/(a^2-bc) +1(/b^2-ca) +(1/c^2-ab)

ab + bc + ca = 0

so bc = - ab - ca = -a(b+c)

so a^2 - bc = a(a + b+ c)

so 1/ (a^2 - bc) = 1/(a(a+b+c))

similarly

1/ (b^2 - ca) = 1/(a(a+b+c))

1/ (c^2 - ab) = 1/(a(a+b+c))

adding we get (1/a^2-bc) +(1/b^2-ca) +(1/c^2-ab)= 1/(a+b+c) ( 1/a + 1/b+ 1/c)

= ( 1/a + 1/b+ 1/c)/(a+b+c)
=(bc + ac + ab)/((a+b+c)(abc)) = 0

Q13/104) Let a,b,c be the roots of x3−7x2−6x+5=0. Compute (a+b)(a+c)(b+c).

Let f(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37

Q13/103) Given f(x) = x -3 for x >= 1000 and f(f(x+5)) x < 1000 find f(84)

we evaluate f(999) through f(995)

f(999) = 998
f(998) = 997
f(997) = 998
f(996) = 997
f(995) = 998

now f(85)= f^184 (999) = f^183( 998) notation for f is applied 183 times

applying f twice gives 998 and so on applying 182 time gives 998 and then once more gives 997 which is the ans

for some discussions refer to http://mathhelpboards.com/challenge-questions-puzzles-28/find-f-84-a-6600.html#post30091

Q13/102) Given abc=1 Show that 1/(1+a+b^-1) + 1/(1+b+c^-1) + 1/(1+c+a^-1) = 1

we have 1/(1+ a + 1/b) = b/(b+ab+1) ...1

1/(1+b+ 1/c) = c/(c+bc+1) = c/(c+bc + abc) ( as 1 = abc)
= 1/( 1+ b+ ab) .. (2)

1/( 1+ c + 1/a) = a/( a+ ac + 1) = ab/( ab + abc + b) = ab/( ab+ 1 + b) ... (3)

adding (1) , (2), (3) we get the result

Q13/101) Solve : 36/x^(1/2) + 9/y^(1/2) = 42 – 9 x^(1/2) – y^(1/2)

36/x^(1/2) + 9x^(1/2) + 9 /y^(1/2) + y^(1/2) = 42

Hence

(36/x^(1/2) + 9x^(1/2) – 36) + (9 /y^(1/2) + y^(1/2)- 6) = 0

Hence 9( 2/x^*1/4 – x^(1/4))^2 + (3/y^(1/4)  – y^(1/4))^2 = 0
=> 2/x^(1/4) – x^(1/4) -= 0 or x = 2^2 = 4

And  3/y^(1/4)  – y^(1/4) = 0 or y = 9