2023/003) Prove that there exists 2023 consecutive natural numbers whose sum is a perect square

 Let the 2023 consecutive numbers be from n-1011 to n + 1011

For all to be natual numbers $n>=1012$ 

The sum of them = 2023n

$2023 =17 ^2 * 7$

So if we choose n to be of the form $7m^2$ then the sum becomes a perect square

Not $n >= 1012$ or $7m^2 >=1012$ of $m > 13$

So the 2023 number starting from $7m^2-1011$ where $m > =13$ satisfy the criteria 

2023/002) A room is empty. Each minute a person enters a room or 2 leave. after exactily $3^{1999}$ minutes can the no of persons be $3^{1000} + 2$

Let persons enter A instances and leave B instances

A+B = $3^{1999}\cdots(1)$

A-B = $3^{1000} + 2\cdots(2)$

Subtracting 2nd from the 1st we get

$3B = 3^{1999} - 3^{1000} - 2$

LHS is multiple of 3 but RHS is not so it is not possible  

2023/001) When dividing a polynomial f(x) by $(x-1)^2$ the remainder is x+1. If f(x) is divided by $x^2$ the remainder is 2x+3 . if the remainder when divided by $x^2(x-1)$ is $ax^2+bx+c$ then find a+b+c

 Dividing a polynomial f(x) by $(x-1)^2$ the remainder is $g(x) = x+1$-

so dividing by (x-1) the remainder is $g(1) = 1 + 1 = 2$

Dividing by $x^2(x-1) $ is $ax^2+bx+c$

so deviding  $ax^2+bx+c$ by $x-1$ remainder must be 2

so $f(1) = a + b+ c = 2$