2018/019) Find the number of pairs (a,b) such that $(a+ib)^{2002} = (a-ib)$

We have
$(a+ib)^{2002} = (a-ib)$

Multiply by (a-ib) on both sides to get

$(a+ib)^{2003} = (a-ib)(a+ib) = a^2+b^2\cdots(1)$
Now from the given condition we have taking mod on both sides
$(\sqrt{a^2+b^2})^{2002} = \sqrt{a^2+b^2}$
Or  $(\sqrt{a^2+b^2})((\sqrt{a^2+b^2}) ^{2002} -1)= 0$
So $(\sqrt{a^2+b^2})= 0\cdots(2)$ or $(\sqrt{a^2+b^2}) = 1\cdots(3)$
From (2)   one gives $(a,b) = (0,0)$ one solution
from (1) and (3) we get
$(a+ib)^{2003} = 1$ and this has got 2003 solutions giving total 2004 solutions

or

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 = $1 + \frac{105-1}{2} = 53$

2018/011) Let $P(n) = (n+1)(n+(n+3)(n+5)(n+7)(n+9)$ What is the largest integer that is a divisor of P(n) for n even

Choose 2 values quite far off (as, if we chose 2 consecutive values then 4 terms shall be same
We have $P(2) = 3 * 5 * 7 * 9 *11$
and $P(10) = 11 * 13 * 15 * 17 * 19$
$GCD(P(2),P(10) = 15$
So GCD of the numbers shall be factor of 15
Now we have $(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)(n+2)(n+1)(n+3) \pmod 3$
and it is product of 3 consecutive numbers we have this is divisible by 3
Now we have $(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)(n+5)(n+2)(n+4) \pmod 5$
and it is product of 5 consecutive numbers we have this is divisible by 5
So it is divisible by 15
So 15 is the required number

2018/010) Find solution of $(2x+1)(30x+1)(3x+1)(5x+1) = 10$

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) = $

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  $x = \dfrac{-4\pm\sqrt{31}}{15}$

2018/009) Solve in integers $x^2+615 = 2^n$

working in modulo 3 we have
$  615 \equiv 0 \pmod 3$
as LHS is 0/1 mod 3 so we must have RHS 1 mod 3(as it cannot be 0 mod 3)
so n has to be even (as 2^n is 1 mod 3 for n even) so n = 2m
so
$x^2+615 = 2^{2m}$ or $2^{2m}- x^2 = 615$ or $(2^m + x)(2^m - x) = 615$

so we have the following combinations (615,1) , (205,3), (123,5), (41,15), out of which we get solutions $(2^m= 64,x= 59)$, others are not power of 2 so $(x=59, n=12)$ is the solution 

2018/008) Five distinct 2 digit numbers are in GP. Find them

As we have $2^4=16$ and $10 * 16 > 99$ hence the common ratio $< 2$

So let the common ration be $\frac{b}{a}$ where a and b are co-primes

So the numbers are $a^4, a^3b,a^2b^2,a^3b,a^4$ and we should have

$a^4 >=10=>a>= 2$ and $b^4 < 100=> b <=3$ ands hence $a=2,b=3$ and the numbers are$16,24,36,54,81$

2018/007) consider the equation $x^4-18x^3+kx^2+174x - 2015=0$

If the product of 2 roots of eqution is -31 then find the value of k

Solution
Product of 4 roots = -2015
product of 2 roots = -31
so product of other 2 roots = -2015/(-31) =  65
so 2 quadratic factors are $x^2+ax-31$ and $x^2+bx +65$ where a and b are to be determined
so  $x^4-18x^3+kx^2+174x - 2015=(x^2+ax-31)(x^2+bx +65)$
or $ x^4 + (a+b)x^3 +(65-31+ab)x^2 + (65a - 31b)x - 2015=0$
comparing coefficients we get
$a+b= -18$, $(65a-31b= 174$,$k= 34 + ab$
we can solve 1st 2 to get $a= -4, b= -14$ so putting in 3rd we get $k= 34 + ab= 90$   

2018/006) Show that if $a+b$, $c+a$, $b+c$ are in HP $a^2,b^2,c^2$ are in AP

Because $a+b$, $c+a$, $b+c$ are in HP
hence
 $\frac{1}{a+b}$, $\frac{1}{c+a}$, $\frac{1}{b+c}$ are in AP
or
 $\frac{1}{b+c}-\frac{1}{c+a}= \frac{1}{c+a}-\frac{1}{a+b}$
or
$\frac{(c+a)-(b+c)}{(b+c)(c+a)}= \frac{(a+b)-(c+a)}{(c+a)(a+b)}$
or
$\frac{a-b}{(b+c)(c+a)}= \frac{b-c}{(c+a)(a+b)}$
or $(a-b)(a+b)=(b-c)(b+c)$
or $a^2-b^2 = b^2 -c^2$
or $a^2,b^2,c^2$ are in AP

2018/005) Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+1}}\right\rfloor$

we realise that $n(n+1) = (n+\dfrac{1}{2})^2 - \dfrac{1}{4}$
so $\sqrt{n(n+1)}\lt(n+\dfrac{1}{2})$
clearly $n\lt\sqrt{n(n+1)}$
so
we have
$(\sqrt{n} + \sqrt{n+1})^2 = n + n+ 1 + 2 \sqrt{n(n+1)}$
= $2n +1 + 2 \sqrt{n(n+1)}$
$>2n+ 1 + 2 n$ or > $4n+ 1$
and = $2n +1 + 2 \sqrt{n(n+1)}$ <  $2n +1 + 2 (n + \dfrac{1}{2})$
or $< (4n + 2)$
so $(4n+1)\lt(\sqrt{n} + \sqrt{n+1})^2\lt(4n+2)$
because  4n+2 is not a perfect square
we have $\lfloor\sqrt{4n+1}\rfloor = \lfloor\sqrt{4n+2}\rfloor$
and as $\sqrt{n} + \sqrt{n+1})^2$ is between 4n + 1 and 4n +2 we have
$\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor= \lfloor(\sqrt{n} +\sqrt{n+1}\rfloor$

2018/004) Show that product of 4 consecutive natural numbers cannot be a perfect cube

Proof:
Here use use the fact that if p and q are co-primes and pq is a cube then both p and q both are cubes
there are 2 cases;
1) the lowest  number is odd.
the numbers are n, n+1, n+ 2, n+ 3 . and n+2 being odd lowest factor or $n+2 >=3$ hence it is co-prime to rest of the numbers
so n+2 is a cube and n(n+1)(n+3) is a cube
$n(n+1)(n+3) = n^3 + 4n^2 + 3n$
$n(n+1)(n+3) - (n+1)^3 = n^2 -1$
this is zero for n=1 which need to be checked and for n = 1 we have product 24 not a cube
for $n > 1$  $n(n+1)(n+3) > (n+1)^3$
$(n+2)^3 - (n^3 + 4n^2 + 3n)  = 2n^2 + 9n - 8 = 2n^2 + n + 8(n-1) > 0$
so it is between $(n+1)^3$ and$ (n+2)^3$ and not a perfect cube
hence it is not a perfect cube for n odd

Now we see for n even

n+ 1 is odd and we need to show that $n(n+2)(n+3) = n^3 + 5n^2 + 6n$

so  $n(n+2)(n+3) - (n+1)^3  = 2n^2 + 3n-1 >0$

 $(n+2)^3 -  n(n+2)(n+3) = (n+2)((n+2)^2-n(n+3))$
$ = (n+2)(n^2+4n+4 -n ^2- 3n) = (n+2)(n+4) >0 $
so it is between $(n+1)^3$ and$ (n+2)^3$ and not a perfect cube

  Hence it is not a perfect cube for any n

2018/003) For the set of equations $z^x=y^{2x}, 2^z= 2.4^x,x+y+z = 16$ find integral values of x,y,z

we are given
$z^x=y^{2x}\cdots(1)$
$2^z= 2.4^x\cdots(2)$
$x+y+z = 16\cdots(3)$

$z^x=y^{2x}$ is true when $z=y^2$ or $x = 0$
we deal both cases
case 1
$z=y^2\cdots(4)$
the equation (2) gives
$2^z = 2^{2x+1}$
or $z=2x+1$
or $2x = z-1=y^2-1\cdots(5)$
from (3), (2), (5) we get
$y^2-1 +2y + 2y^2 = 32$
or $3y^2 +2y-33=0$ or $(y-3)(3y+11)=0$ so y =3 as y is integer
hence z=9 and x= 4
so solution $x=4,y=3,z=9$
case 2
x = 0
so from (2) $2^z =2=>z=1$
hence y = 15
so solution $x=0,y=15,z=1$

2018/002) Find the GCD of all the even numbers formed from permutation of $1,2,3,4,5,6$

Each of the number is divisible by 6.
Let us take 2 number 123546, 123564
Let us find the GCD of the same
$GCD(123546,123564) = GCD(123546,123464-123546)$
$=GCD(123546,18) = GCD(6863 * 18+12,18) = GCD(12,18) = 6$
hence from above GCD = 6

2018/001) Show that if a,b,c are sides of a triangle then $(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$

We have
$a^2 >= a^2 - (b-c)^2$ or $a^2 >= (a+b-c)(a-b + c) \cdots(1)$
Similarly $b^2 >= (b+c-a)(b-c+a)\cdots(2)$
and  $c^2 >= (c+a-b)(c-a+b)\cdots(3)$

Multiply (1) , (2) , (3) to get
$(abc)^2 >= ((a+b-c) (b+c-a)(c+a-b))^2$
or
$abc > = (a+b-c)(b+c-a)(c+a-b)\cdots(4)$

Applying $AM >= GM$ to $a,b,c$ we get

$\frac{a+b+c}{3} >= \sqrt[3]{abc}$

or $(a+b+c)^3 >= 27abc\cdots(5)$

From (4) and (5) we get

$(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$