2019/018) $x^2 + bx + a$ leaves same remainder when divided by x + 2 or x - a (where a ≠ -2). Show that a + b = 2.

We have
$f(x) = x^2 + bx +a$
Same remainder when divided by x + 2 or (x-a)

So $f(-2) = 4- 2b + a = f(a) =  a^2 +ab +a$
Or 4 - 2b = a^2 + ab
Or 2b+ ab = 4 - a^2
Or b(2+a) = (2-a)(2+a)
As a is not -2 so we have b = 2-a or a + b = 2

Alternatively as same remainder when divided by x+ 2 and x -a and it is degree 2 polynomial

We have $f(x) = x^2 + bx + a = (x+2)(x-a) +c $ where  c is a constant
Or  $f(x) = x^2 + bx + a = x^2+(2-a)x + c - 2a$
Comparing coefficient of x we have b= 2- a or a+b =2 

2019/017) Each of the numbers $x_1,x_2,\cdots,x_{101}$ is $\pm1$. what is the smallest positive value of $\sum_{1 \le x_i\lt x_j\le 101 }x_ix_j$

Let the given sum be S

We have $(\sum_{1 \le  n \le 101 }x_n)^2= \sum_{1 \le  n \le 101 }x_n^2 + 2 \sum_{1 \le  x_i\lt x_j\le 101 }x_ix_j  $

so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - \sum_{1 \le  n \le 101 }x_n^2$

 as each $x_i$ is $pm1$ so $x_i^2 = 1$

so $\sum_{1 \le  n \le 101 }x_n^2 = 101$
so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - 101$
 For S to be positive we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 \gt   101$ and odd
for S to be smallest we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 $ smallest number greater than 101 and it is true when $(\sum_{1 \le  n \le 101 }x_n)^2 = 121$
this is possible and this give S = 10.

2019/016) Let the rational number $\frac{p}{q}$ be closest to but not equal to$\frac{22}{7}$ among all rational numbers with denominator less than 100. What is the value of $p-3q$

Solution

We have $\lvert \frac{p}{q}-\frac{22}{7}\rvert$ as close to zero
Or $\lvert \frac{7 p- 22q}{7q}\rvert$
Above value is lowest when 7q is close to the highest and 7p-22q is close to the lowest( as both may not be at the same point)
q highest is when q = 99  this gives p = 311 and 7p-22q = 1 so this is lowest when q is highest
so p = 311, q = 99 and p - 3q = 311 - 3 * 99 = 14

2019/015) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^ab^b +a^bb^a <=1$

without loss of generality we can assume $a>=b$
We have
$1= a+ b = a^{a+b} + b^{a+b}$
so $1- (a^ab^b + a^b b^a)$
$=  a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$
$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$
for a > b both the terms are non -ve so we have

$1- (a^ab^b + a^b b^a) >=0$ and hence the result

2019/014) Find the values of n such that $n^4+4$ is a prime

We have $n^4+4 = n^4+4n^2 + 4 - 4n^2 = (n^2+2)^2 -(2n)^2 = (n^2+2n+2)(n^2-2n +2)$
$n^4+4$ is a prime iff $n^2+2n+2$ is a prime and $n^2-2n+2=1$
$n^2-2n+2=1=>(n-1)^2 = 0$  or n = 1
And for n= 1 $n^2+2n+2=5$ which is a prime
we could also compute $n^4+4= 1 + 4 =5$
So 1 is the only choice  for n

2019/013) The length of perimeter of a right $\triangle$ is 60 inches and the length of altitude perpendicular to the hypotenuse is 12 inches. find the sides of the $\triangle$

Let the sides be a,b,c where c is the hypotenuse. without loss of generality we assume $a>=b$

We have
$a+b+c= 60$
Or $60-c = a+b\cdots(1)$
$a^2 + b^2 = c^2\cdots(2)$
Area of the triangle $\frac{1}{2}ab = \frac{1}{2}12c$ or
$ab= 12c\cdots(3)$
From (1)
$(60-c)^2 = ( a + b)^2 = a^2 + 2ab + b^2 = (a^2+b^2) + 2 * 12c= c^2 + 24c$ (using (2) and (3))
Or $3600 - 120c + c^2  = c^2 + 24c$
Or $144c = 3600$ or $c=25$
From (1) $a+b= 35\cdots(4)$
From (3) $ab= 300\cdots(5)$
So we have $(a-b)^2 = (a+b)^2 - 4ab = 1225 - 1200 = 25$ (using (4) and (5))
So $a-b = 5\cdots(6)$
From (4) and (6) we have a = 20, b = 15,
So sides of triangle $15,20,25$