2025/029) Prove that if $x^2+xy+y^2$ is divisible by 10 it is divisible by 100

As 10 = 2 *5 let us check for both 2 and 5

We have $x^2+ xy+y^2 = x^2+ y(x+y)$

if x is odd the $x^2$ is odd and either y or x+y is even so the expression is odd 

similarly if y is odd.

So both x and y are even  and each term is divisible by 4 so the sum

let the expression be divisible by 5

working in mod 5 let $x=0$ we get $x^2+xy + y^2= y^2$ and it is divisible by 5 only if $y=5$ then each term is divisible by 25 so the expression.

Now let us have x is not zero .let $y = mx$

so we get $x^2+xy + y^2=  x^2(1+m+m^2)$

as x is non zero deviding by x we get $1+m+m^2$ ant for m = 0 to 4 we get it not a multiple of 5 so x and y has to be zero mod 5

So we have the expression divisible by 25 and also 4 so by 100

Proved  

 

 

 

2025/028) How do you prove that $\cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ =\frac{1}{16}$

Let $A= \cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

We have as 80 is double of 40 and 40 is double of 20 let us multiply both sides by $2\sin\,20^\circ$ we get 

  $A*2\sin20^\circ = 2\sin20^\circ\cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

or

$A*2\sin20^\circ = \sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$ 

or  multiplying by 2 we get 

 $A*4\sin20^\circ = 2\sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

or 

$A*4\sin20^\circ = \sin80^\circ \cos 60^\circ \cos 80^\circ$ 

 or $A*8\sin20^\circ = 2\sin80^\circ \cos 60^\circ \cos 80^\circ$

  or $A*8\sin20^\circ = 2\sin80^\circ \cos 80^\circ \cos 60^\circ$

 or $A*8\sin20^\circ = \sin160^\circ \frac{1}{2}$

 or $A*8\sin20^\circ = \sin20^\circ \frac{1}{2}$ (as $\sin20^\circ = \sin 160^\circ$)

or $A * 8 = \frac{1}{2}$

or $ A= \frac{1}{16}$  

Hence proved  

 

2025/027) Let a,b,c be the lengths of the sides of a triangle. Show that if $a^2+b^2+c^2=ab+bc+ca$ then the triangle is equilateral.

 we have $a^2+b^2+c^2= ab + bc + ca$

Hence $a^2+b^2+c^2 -ab - bc -ca = 0$

Hence $2a62+2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0$

Or $(a^2-2ab + b^2) + (b^2-2bc+c^2) + (c^2-2ac+a^2) = 0$

or $(a-b)^2 + (b-c)^ + (c-a)^2 = 0$

as each term is non -ve so each of then has to be zeo so a= b= c and hence the triange is equilateral 

2025/026) prove $a(b-c)^2 +b(c-a)^2+c(a-b)^2+8abc=$ $(a+b) (b+c) (c+a)$

 We have as on RHS (a+b) we can add 4abc to the 3rd term to get (a+b) as a factor and add 2abc to 1st and 2nd term that is distributing 8abc to get

 $a(b-c)^2 +b(c-a)^2+c(a-b)^2+8abc$ 

$=a((b-c)^2 +2bc)+b((c-a)^2+2ac)+c((a-b)^2+4ab)$

 $=a((b^2+c^2)+b((c^2+a^2))+c(a+b)^2$

 $=ab^2+ac^2+bc^2+ba^2+c(a+b)^2$

  $=ab^2+ba^2+ac^2+bc^2+c(a+b)^2$

$=ab(a+b) + c^2(a+b) + c(a+b)^2$

$=(a+b)(ab+c^2 + c(a+b)$

$=(a+b) (c^2 + ca + bc + ab)$

$= (a+b)(b+c)(c+a)$ 

 

2025/025) How many numbers (n) are there between 1 and 200 such that $\frac{n}{2}$ , $\frac{n}{3}$ ,$\frac{2n+1}{5}$ are all composite natural numbers (CAT 2020)?

First let us find the number such that $\frac{n}{2}$ ,  $\frac{n}{3}$ ,$\frac{2n+1}{5}$ are integers

as  $\frac{n}{2}$ ,  $\frac{n}{3}$ are integers so $\frac{n}{6}$ is integer say k

so n= 6k

As in denominator we have 2 3 and 5 so we should have mod 30

take k from 0 to 5 we get n = 12 satisfies  $\frac{2n+1}{5}$ integer

so we get values n =12 , 42, 72, 102,132,162,192 .

as n is multiple of 6 $\frac{n}{2}$ ,  $\frac{n}{3}$ are composite we need to check   $\frac{2n+1}{5}$ composite . let us compute $\frac{2n+1}{5}$

n = 12 gives 5 so no

n = 42 gives 17 so no

n = 72 gives 29 no

n =  102 gives 41 no

n = 132 gives 53 no

 n = 162 gives 65 yes

n = 192 gives 77 yes

so there are 2 values  

 

 

 

2025/024) show that $3^{(2n + 1)}+ 2^{(n + 2)}$ is dvisible by 7

we have 

 $3^{(2n + 1)}+ 2^{(n + 2)}$

 $=3 *3^{2n}+ 4 *2^{n}$ 

 $=3 *3^{2n}+ 4 *2^{n}$

 $=3 *9^{n}+ 4 *2^{n}$  

 $=3 *2^{n}+ 4 *2^{n}$  as $9 \equiv 2 \pmod 7$

 $=(3 + 4) *2^{n}$  

  $= 7 *2^{n}$  divisible by 7 

2025/023) Solve $\sqrt{(x+2)} > x $

It is risky to square both sides as it shall give erroneous root.

Let us first find the range of x

as we can take the square root of a non -ve number only so we 

$x \ge -2$ 

for $x \in [-2,0] LHS is positve and RHS is -ve or zero. so  this is range in  -ve

To find the range is positive we square both sides to get 

$(x+2) > x^2$

or   $x^2-x-2 <0$

or $(x-2)(x+1) <0$ or   $x \in [-1,2)$

as x is positive so   $x \in [0,2)$

so combining we get  $x \in [-2,2)$