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  $x \in [-2,2)$ 

 

2025/022) How do you prove that $a^5−a$ is divisible by 10 (a is a natural number)

 Both a and $a^5$ are even or odd so $a^5-a$

Now if a is multiple of 5 $a^5$ is multiple of 5. so $a^5-a$ is divisible by 5

If a is not multiple of 5 then as per FLT $a^5 \equiv a \pmod 5$ so  $a^5-a$ is divisible by 5

In both cases  $a^5−a$ is divisible by 10

 

 

 

2025/021) Find integer solutions of the equation $2x - 3y = 8$

we can solve by fining GCD of 2 and 3 but here I shall solve it by a different approach 

We can rewrite this as

$3y = 2x -8$

RHS is even so LHS should be even or y should be even

So let $y =2k$

We get $6k = 2x -8$

Or x = $3k + 4$

So $x = 3k + 4$ and $y = 2k$ are parametric form of solution

by choosing any integer for k we can get particular solution

k = 1 gives (7,2)

k =2 gives (10,4)

2025/020) Given $ab = 10$ and $a- b= 3$ find $a^3-b^3$

We can solve the 2 equations and get $a=5$ and $b=2$ and hence $a^3-b^3=5^3-2^3=117$

but the above approach is longer as compared to the approach below

We have

$(a-b)^3 = a^3-3a^2b + 3ab^3 - b^3 = a^3-b^3-3ab(a-b)$

or $a^3-b^3= (a-b)^3 + 3ab(a-b)$

Putting the values we get $a^3-b^3=3^3 + 3 * 10 * 3 = 117$

In this process we have avoided solving of equations