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