Saturday, September 23, 2017

2017/025) Solve $2^y+1 = x^2$

We have
$2^y = (x^2-1)= (x+1)(x-1)$
both x+ 1 and x-1 must be power of 2. difference of them is 2.
2 powers of 2 that differ by 2 are 2 and 4

so x+ 1= 4 and x-1 = 2 or x =3 and y = 3

Tuesday, September 19, 2017

2017/024)If m is a root of $A(x)=x^2−nx+10=0$, and n is a root of $B(x)=x−mx+20=0$, what is the value of $(\frac{n}{m} + \frac{m}{n} )^2$

m is root of  $A(x)=x^2−nx+10=0$
so $m^2-mn+10= 0$

n is root of $B(x)=x−mx+20=0$ hence  $n^2-mn+20 = 0$

so we have $m(m-n) = - 10$ and $n(m-n) = 20$ or n = -2m

so $\frac{n}{m} = -2$ and $\frac{m}{n} = -\frac{1}{2}$ or $(\frac{n}{m} + \frac{m}{n})^2 = \frac{25}{4}$

2017/023) Find natural number n such that $n^4+33$ is a a perfect square

we need to have $n^4+33 >= (n^2+1)^2$ as next of $x^2$ is $(x+1)^2$
so $33 >= 2n^2 +1$ or $n^2 <=16$ or $n<=4$
n cannot be odd as $n^4+33$ shall be 2 mod 4 and cannot be perfect square
so n = 2 or 4
both are solutions as $2^4+33= 49 = 7^2$ and $4^4+33=289=17^2$


Saturday, September 2, 2017

2017/022) Solve $(2x+1)(30x+1)(3x+1)(5x+1) = 10$

we have $(2x+1)(30x+1)(3x+1)(5x+1) = 10$
or
$(60x^2+ 32x + 1)(15x^2+ 8x + 1) = 10$

letting $15x^2 + 8x = t$
$(4t+1)(t+1) = 10$
or
$4t^2 + 5 t + 1 = 10$
or $4t^2 + 5t - 9 = 0$
or $(4t+9)(t-1) = 0 $

t =  1 or -9/4
t = 1 gives
$15x^2 + 8x-1=0$ giving $x = \dfrac{-4\pm\sqrt{31}}{15}$

or  $(15x^2 + 8x +\frac{9}4{4}) = 0$
or $(60x^2+ 32x + 9) = 0$
this gives complex solution
so solutions are  $x = \dfrac{-4\pm\sqrt{31}}{15}$

2017/021) find integer n where $\dfrac{99^n+19^n}{n!}$ is highest

Let us consider $\dfrac{99^x}{x!}$

this value increases for x upto 98 and at 99 the value is same as at 98.

now for the second part $\dfrac{19^x}{x!}$ the value increases for x upto 18 and at x = 19 is same as at x = 18 and then it decreases

for x = 19 to 98 the term $\dfrac{99^x}{x!}$ increases more rapidly as $\dfrac{19^x}{x!}$ decreases
at x = 19 the 1st term is higher by $3.8^{19}$ times and it is much higher
so we need to compare the value at x = 98 and x = 99.
the second term  is fighter at x = 98

so $\dfrac{99^x+19^x}{x!}$ is highest at x = 98