2020/018) Prove that $3^n>=27n^3 $ for $n >=9$

To prove the same we use principle of mathematical induction

Base step

For n = 9 $LHS = 3^9 = 3^3 * 3^6 = 27 * 9^3$ so $3^n >= 27n^3$

SO base step is true

Now $(\frac{n+1}{n})^3$ decreases as n increases and at n = 9 we have   $(\frac{n+1}{n})^3= \frac{1000}{729}< 3$

So $(\frac{n+1}{n})^3< 3$ for all $n>=9$

Or $3 > (\frac{k+1}{k})^3\cdots(1)$ for all $k>=9$

Let it be true for n = k $k >=9$

We need to prove it to be true for n =  k+ 1

$3^k > = 27 k^3$

Multiplying by (1) on both sides

$3^{k+1} > = 27 * (\frac{k+1}{k})^3 * k^3 $

Or $3^{k+1} >= 27(k+1)^3$

So it is true for n = k+ 1

We have proved the induction step also

Hence proved

2020/017) FInd q when the equation $x^4-40x^2+q=0$ has 4 roots is AP.

Because it has 4 roots in AP so let the roots be $a-3d, a-d,a+d, a+3d$ 
The sum of the root is zero as coefficient of $x^3=0$
So we have $(a-3d)+(a-d)+(a+d)+(a-3d) = 4a = 0$
Or $a=0$
Hence the roots are $-3d, -d, d, 3d$
So Equation becomes  $(x+3d)(x+d)(x-d)(x-3d)=0$
Or $(x+3d)(x-3d)(x+d)(x-d)=0$
Or $(x^2-9d^2)(x^2-d^2) =  x^4-10d^2x^2+ 9d^4=0$
Comparing with given equation $-10d^2= - 40$ or $d^2=4$
And $q=9d^4= 9 (4)^2= 144$
Hence $q= 144$

2020/016) find n such that $\sqrt{n} + \sqrt{n+2005}$ is a natural number

Let $\sqrt{n+2005} + \sqrt{n} =m \cdots(1)$

We know $(n+2005) - (n) = 2005$
Or $(\sqrt{n+2005})^2 - (\sqrt{n})^2 = 2005$
Or $(\sqrt{n+2005} +  \sqrt{n}) (\sqrt{n+2005} - \sqrt{n}) = 2005\cdots(2)$

Dividing (2) by (1) we get
$(\sqrt{n+2005} - \sqrt{n}) = \frac{2005}{m}\cdots(3)$

Subtracting  (3) from (1) we get $2\sqrt{n}= m - \frac{2005}{m}$

Clearly we have $m^2>=2005$ so we choose m factor of 2005 that is 401, 2005

Taking $m= 2005$ we get $n = (\frac{1}{2}(2005-1)^2 = (1002)^2 = 1004004$

Taking $m= 401$ we get $n = (\frac{1}{2}(401-5)^2 = (198)^2 = 39204$

So we have solution set $(1004004,39204)$

2020/015) Find integers n and k such that $n!+8 = 2^k$

Firstly $n\lt 6$ because if $n >= 6$ then is is divisible by $2^4$ ( 2 comes once in 2 twice in 4 and once is $6$ so at least 4 times and $2^4=16$.

we can show it in another way that $6!=720 = 16 * 45$
so $n>6$ $n!$ shall be divisible by 16 so say value is 16m

now $n!+8 = 16m+8 = 8(2m+1)$ and as it is product of 8 and an odd number greater than 1 so it cannot be power or 2 .

Further n cannot be less than 4 because 8 has to  to factor of n

so we need to check for n = 4 and n = 5

no $4!+8 = 24 + 8 = 32$ giving n =  4 k = 5

$5! + 8 = 120 + 8 = 128 = 2^7$ giving n =  5 and k = 7

Q2020/14) Solve in natural numbers $\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}$

we get
$2018b + 2018 a = 3ab$
or $3ab - 2018 a - 2018b = 0$
As the coefficient of a,b are same so we get product of 2 monomials one of a one of b and coefficients being same.
Multiplying by 3 we get
$9ab - 3 * 2018 a - 3 * 2018b = 0$
or $9ab - 3 * 2018 a - 3 * 2018b = 2018 ^2 $
or $(3a-2018)(3b-2018) = 2018^2= 2^2 * 1009^2$
LHS each term is 1 mod 3 so we should make the RHS so product of 2 numbers each of which is 1 mod 3.


if(a,b) is a solution then (b,a) is also a solution so let us assume that $a>=b$

so we get following sets

$3a-2018= 2018^2, 3b-2018=1$  giving $(a= 1358114,b= 673)$
$3a-2018= 1009^2 , 3b=2018=4$ giving $(a= 340033, b= 674)$
$3a-2018= 1009, 3b-2018=1009 * 4$ giving $(a= 1009,b= 2018)$

and permutation of the same.

2020/13) Solve $2\log_2(x+15) - \log_2x = 6$

We have
$2\log_2(x+15) - \log_2x = \log_2(x+15)^2 - \log_2(x) = \log_2\frac{(x+15)^2}{x} = 6$
Or $\frac{(x+15)^2}{x} = 2^6= 64$
Or $(x+15)^2 = 64x$
Or $x^2+30x+225-64x=0$
Or $x^2-34x+ 225=0$
Or $(x-25)(x-9)= 0$ So x = 25 or 9 

2020/012) Solve in real $x^4+y^4+z^4 + 1 = xyz$

We have LHS is positive so RHS is also positive.

So all of x,y,z are positive or one is positive and rest 2 are negative

So let us solve for positive x,y,z

Using AM GM inequality we have  $\frac{x^4+y^4+z^4+1}{4} >=\sqrt[4]{x^4y^4z^4}$

Or $x^4+y^4+z^4 +1 >= 4xyz$

And they are equal if $x=y=z$

So we get $3x^4-4x^3+1 = 0$

Or  we have trying $x=1$ and $x = 3$ x=1 is solution

giving one set of solution = $(1,1,1)$ and 2 negative  and one positive gives 3 more $(1,-1,-1), (-1,-1,1), \,and\, (-1,1,-1)$