2024/048) Show that if p is prime then ${2p}\choose{p} $$ \equiv 2 \pmod p$

 We have

 ${2p}\choose{p}$$=\frac{(2p)!}{p!p!}$ bu definition

$=\frac{p!\prod_{k=p+1}^{2p}k}{p!p!}$ bu expansion

$=\frac{\prod_{k=p+1}^{2p}k}{p!}$ cancelling p! from both numerator and denominator

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!}$ by taking 2 p ouut

working in mod p we get $(p+n) \equiv n \pmod p$

So we get

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!} \pmod p$

$=\frac{2p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p!}{p!} \pmod p$

$=2$ 

Proved 

2024/047) Express the number 2024 as the sum of some positive integers in such a way that the product of these positive integers is maximal.

When we  break a number into smaller parts the product become larger as number of parts becomes larger and each is smaller however no part can be 1 as it shall provide a smaller value. so we can break it into 1012 parts and each is 2 giving a product $2^{1012}$.

However there is exception to ir as $2^3 \lt 3^2$ so we can group terms of 3 getting 674 3 and left with 1 2 giving a product $3^{674} * 2$

2024/046) Show that cube of a positive integer can be written as difference of 2 squares

 Proof:

A cube is either even or odd

If it is odd we can write it as $2n + 1$ which can be written as $(n+1)^2 - n^2$

If it is even the the number is even say $2n$ so cube is $8k$ where $n= k^3$  can be written as $(2k+1)^2 -(2k-1)^2$

2024/045) Solve in positive integer $\frac{x}{y+7} + \frac{y}{x+7}= 1$

We are given  $\frac{x}{y+7} + \frac{y}{x+7}= 1$

or $x(x+7) + y(y+7) = (x+7)(y+7)$

or $x^2+7x + y^2 +7y = xy + 7x + 7y + 49$

or $x^2 + y^2 -xy = 49$

now $x^2- xy$ we need to form in form of squares by adding some teerm

let us multiply by 4 to above to get

    $4x^2 +  - 4xy + 4y^2 = 196$

or $4(x^2-4xy + y^2) +3y ^2 = 196$

or $(2x-y)^2 + 3y^2 = 196$

as it is sum of squares  we need to check a finite number of values(fron pair of values we consider only positive ones) 

$y = 3$ gives $2x-y =13$ giving $x = 8, y = 3$

$y= 5$ gives $2x-y = 11$ giving$ x = 8, y = 5$

$y = 7$ gives $2x -y = 7$ giving $x = y = 7$

$y = 8$ gives $2x -y = 2$ giving $x = 5 y = 8$

         or $2x - y = -2$ giving $x = 3 and y = 8$

so the solutions are $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$

 

2024/044) What is the equation of a circle which passes through three points $(0,0), (a,0),(0,b)$

 https://www.quora.com/What-is-the-equation-of-a-circle-which-passes-through-three-points-0-0-a-0-and-0-b

Because one point $(0,0)$  is origin, one point $(a,0)$ lies on x-axis and another $(0,b)$ on y axis it is clear that is is a right angled triangle. so the circumcenter is the midpoint of the hypotenuse that is $(\frac{a}{2},\frac{b}{2})$

so the equation of the circle is

$(x-\frac{a}{2})^2 +(y-\frac{b}{2})^2=(0-\frac{a}{2})^2 + (0-\frac{b}{2})^2 $  the RHS is square of  distance of origin to to the centre

simplifying we get

$x^2-ax+y^2-by=0$

 

2024/043) Given $x^2 = 2^ y + 2023$ find x+ y where x and y are natual numbers(IIT JEE 2024)

We are given

  $x^2 = 2^ y + 2023$

Now working mod 4 we have 

$2^ y + 2023 \equiv 2^y + 3 \pmod 4$

y cannot be greater than 1 as $2^y + 3 \equiv 3 \pmod 4$

As a square cannot be $3 \pmod 4$ so only possible value is y = 1 giving x= 45 and $x+y=46$

2024/042) For each positive integer n we consider the sequence of 2004 integers $\lfloor n+\sqrt{n}\rfloor ,\lfloor n+1+\sqrt{n+1}\rfloor ,\lfloor n+2+\sqrt{n+2}\rfloor,$ $\cdots,\lfloor n+2003+\sqrt{n+2003}\rfloor$. How do I find the smallest integer n, such that the 2004 numbers in the sequence are 2004 consecutive integers?

We know for any integer x

$\lfloor x+y \rfloor = x + \lfloor y \rfloor$

So we have

$\lfloor  n + k + \sqrt{n+k}  \rfloor = n + k  + \lfloor \sqrt{n+k} \rfloor$

From the given 2024 numbers by putting k = 0 t0 2023 we get

 $\lfloor  n +  \sqrt{n}  \rfloor = n +  \lfloor \sqrt{n} \rfloor$

 $\lfloor  n + 1 + \sqrt{n+1}  \rfloor = n + 1  + \lfloor \sqrt{n+1} \rfloor$

$\lfloor  n + 2 + \sqrt{n+2}  \rfloor = n + 2  + \lfloor \sqrt{n+2} \rfloor$

$\lfloor  n + 2023 + \sqrt{n+2023}  \rfloor = n + 2023  + \lfloor \sqrt{n+2023} \rfloor$

They are consecutive if we have

 $ \lfloor \sqrt{n} \rfloor = \lfloor sqrt{n+1} \rfloor \cdots \lfloor \sqrt{n+ 2023} \rfloor$

as these are in increasing order we have

  $ \lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+ 2023} \rfloor$

The 1st term should be as low as possible so n is a perfect square $x^2$

So we have $ x = \lfloor \sqrt{x^2+ 2023} \rfloor$

 As we have the smallest $k=(x+1)^2$ such that $ x <  \sqrt{k}$

so we have $x^2 + 2023 \lt x^2 + 2x + 1$ or $2024 \lt 2x $ so x = 1013 and

smallest $n = 1013^2= = 1026169$