2018/018) Find all natural numbers n that satisfy $1 + \lfloor \sqrt{2n} \rfloor | 2n$

Let $x =  \lfloor \sqrt{2n} \rfloor$
Hence $x^2 <= 2n < (x+1)^2$
Or $x^2 <= 2n < x^2+2x +1$
Or $2n = x ^2 + p$ where $0 <= p < 2x +1\cdots(1)$
Now from the given condition
$1+ x | x^2 + p$ as $2n = x^2 + p\cdots(2)$
Or $1 + x | (x^1-1) + (p+1)$
And as $1+ x | (x^2-1)$
We have $ 1 + x | p+1$
x cannot be zero as x=0 gives n = 0 which is not natural number
Let $p+1 = m(1+x)$
As $p + 1 < 2x+2$
So $m(1+x) < 2x +2$ from (1)
Or $m < 2$
And m is not -ve
As we have $m=0$ or $m = 1$
m=0 gives p = -1 which is not admissible
m= 1 gives $p =  x$ or $n= \frac{x(x+1)}{2}$

so n is number of the form  $n= \frac{x(x+1)}{2}$

2018/017) Solve for non -ve integers x, y $y^3=x^3+8x^2-6x+8$

We have
$y^3-x^3= 8x^2-6x+ 8>=8$ for $x>=0$
And $(x+3)^3 = x^3 + 9x^2 + 27x + 27 >y^3$
So we need to consider $y=x+1$ and  and $y=x+2$
Putting $y=x+1$ we get $x^3+3x^2+ 3x + 1 = x^3+8x^2-6x+8$
Or $5x^2-9x+7=0$
This equation does not have any real solution
Putting $y=x+2$ we get $x^3+6x^2+ 12x + 8 = x^3+8x^2-6x+8$
Or $2x^2-18x=0$
$=>x(x-9)=0$
Giving x = 0 or 9
That is solution set $(0,2)$ and $(9,11)$

2018/016) For what natural numbers can the product of some numbers of $n,n+1,n+2,n+3,n+4,n+5$ be same as product of other numbers

As it is sequence of 6 consecutive numbers more than one number cannot be divisible by 7.

If one number is divisible by 7 then it cannot be divided to 2 groups for product to be same.

So the numbers have to be  of the form 7m+1,7m+2,7m+3,7m+4,7m+5,7m+6 and product of them mod 7 is 6. and hence it is not a square ( square mod 7 are 1,4,2).

So no solution exists

2018/015) Find all positive n such that $3^{n-1} +5^{n-1} $ divides $3^n + 5^n$

We have $3^n + 5^n= 3(3^{n-1} +5^{n-1}) + 2*5^{n-1}$
So if  $3^{n-1} +5^{n-1} $ divides $3^n +5^n $ then it divides $2*5^{n-1}$
But $3^{n-1} +5^{n-1} $ does not divide $5^{n-1}$ and they are co-primes
So $3^{n-1} +5^{n-1} $ divides 2
So $3^{n-1} +5^{n-1}$ = 1 or 2 so we get n= 1

2018/014) Find all triples (p,x,y) such that $p^x= y^4+4$ where p is a prime and x and y are natural numbers

We have $p^x=(y^4+4) = (y^4+ 4y^2+4) - 4y^2= (y^2+2)^2-(2y)^2$
Or $p^x= (y^2+2y+2)(y^2-2y+2)$
Now as p is prime we have both $y^2+2y+2)$ and $(y^2-2y+2)$ are powers of p and as
$(y^2-2y+2) < (y^2+2y+2)$
So $(y^2-2y+2)$ divides $(y^2+2y+2)$ divides the difference that is 4y
So $y^2 - 2y +2 -4y <=0$ or $y^2-6y+2<=0$ or $(y-3)^2< 7$ or $ y - 3 < 3$ or $y < 6$
Putting y = 1 to 5 in original equation we see that (p,x,y) = (5,1,1) is the only solution

2018/013) Find the function $f(x)$ if: $f''(x)=12x+8$ and $f'(-1)=0$ $f(-1)=0$

$f''(x) = 12 x +8$ 
so integrate to get 
$f'(x) = 6x^2+8x + C$ where C is constant of integration 
$f'(-1) = 6 = 8 + C = 0$ or C = 2 
so f$'(x) = 6x^2+ 8x + 2$ 
integrate once more 
$f(x) = 2x^3 + 4x^2 + 2x + D$ 
$f(-1) = -2 + 4 - 2 + D = 0$ or D = 0 
so $f(x) = 2x^3+ 4x^2 + 2$

2018/012) How many pairs of numbers are there whose LCM = 600

We have $600= 2^3 * 3 * 5^2$.
For 2 number to have LCM they have to be of the form $2^a3^b4^c$ so let one number
be $2^r3^s5^t$ and aother number be $2^m3^n5^p$. and
$0 <= r < 3$, $0 <= m <=3$ and either r or m = 3
So we have 7 choices for r,m r=3, m= 0 to 3 ( 4 choices) or m =3 r = 0 to 2 ( 3 choices) (m =3, r =3
is already taken care of)
$0 <= s < 1$, $0 <= n <=1$ and either s or n = 1,
3 choices s=0 , n= 1 or s =1 , n = 1 or s =1 , n = 0
$0 <= t < 2$, $0 <= p <=2$ and either t= 2 or p= 2
So we have 5 choices for t,p  t = 2, p = 0 to 2 ( 3 choices) or p =2 t = 0 to 1 ( 2 choices)
So number of pairs = 7 * 3 * 5= 105
Out of which both are 600 is one case.
(m, n) pair is same as (n,m)
So total number of ordered pairs = $1 + \frac{105-1}{2} = 53$