2021/101) Show that any number > 6 can be written as sum of 2 co-prime numbers

Let us consider 3 cases. Let the number be n 

1) n is odd then 2 numbers  are (2, n-1)

2) n is even and of the form 4n. Then the 2 numbers (2n-1, 2n+1)

3) n is even and of the form 4n +2. Then the 2 numbers (2n -1 , 2n + 3)  

2021/100) prove: $\sum_{k=0}^{n} \binom{n}{k}^2 =\dfrac{(2n)!}{(n!)^2}$

Out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

Now let us make 2n objects into 2 groups of n objects each.

For picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

The 2 above are same as it shows the number of ways in 2 different ways

So

$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

Now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result

2021/099) Find the minimum value of $\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

 We have 

$\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

= $\sqrt{​(x+​2)^2+​9}+​\sqrt{​(x-4)^2 + 25}$

The 1st term that is $\sqrt{​(x+​2)^2+​9}$ is distance from (x,0) to (-2,-3) and second term $\sqrt{​(x-4)^2+25}$ is distance from (x,0) to (4,-5).

clearly sum of the distanace to (x,0) is lowest when (x,0), (-2,3) and (4,5) are in a straight line that is (x,0) is on the line from (-2,3) to (4,5)  now as (x,0) lies in between the minimum is that distance from (-2.-3) to (4,5) or $\sqrt{(4+2)^2 + (5+3)^2} = \sqrt{100} = 10$ 

2021/098) Find integers x,y z, such that $x + y +z = xyz$

 Because of symmetry if (x,y,z) is a solution then any permutation of (x,y,z) is also a solution

without loss of generality let us assume that $ x<=y<=z$ 

So $x+y+z <= 3z$ 

puttying in the given equation we get

$3z >= xyz$

or $3>=xy$

This gives the following set in (x,y)

(1,1) giving 2 + z = z which does not have a solution

(1,2) giving 3 + z = 2z or z = 3

(1,3) giving 4 + z = 3z giving z = 2 which is a contradiction as it should not be less than 3

so solution set (1,2,3) or a permutation of the same.

2021/097) Solve the following equation: $[x]^2=[2x]-1$ where [x] is the floor value of the x real No

Let x = n + r where n is the integer part and r is the fractional part

we have

$\lfloor x \rfloor ^2 = 2(x+r) - 1$

so $n^2 = 2n -1$ when $ r < \frac{1}{2}$ or $n^2 = 2n$  and $ \frac{1}{2} \le   r <  1$

$n^2 = 2n -1$ when $ r < \frac{1}{2}$

gives $n^2 - 2n + 1 = 0$ or $(n-1)^2 =0 $ or n= 1 giving $ 1 \le x < 1.5$

$n^2 = 2n$  and $\frac{1}{2} \le   r <  1$

gives n = 0 or 2  giving $ .5 \le x < 1$ or  giving $ 2.5 \le x < 3$

combining them we have  $ .5 \le x < 1. 5 $ or  $ 2.5 \le x < 3$

 

2021/096) Given an arithmetic series a with a common difference d $a_1 + a_{2n+1}= z$ and $\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$ $\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$ Show $d = \frac{y-x}{2nz}$

We are given

$\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$

$\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$

subtract (1) from (2) to get

$\sum_{k=1}^{n} (a_{2k}^2- a_{2k-1}^2)  = y-x$

Or $\sum_{k=1}^{n} (a_{2k}- a_{2k-1})(a_{2k} + a_{2k-1})   = y-x$

But $(a_{2k}- a_{2k-1}= d$ common difference so we get

$\sum_{k=1}^{n} d(a_{2k} + a_{2k-1})   = y-x$

or $d \sum_{k=1}^{n} (a_{2k} + a_{2k-1})   = y-x$

Or $d \sum_{k=1}^{2n} (a_{k})   = y-x\cdots($

as $a_k = a_1 + (k-1) d$ for any k so we have

Now $a_k + a_{2n+1-k} = a_1 + (k-1)d + a_1 + (2n+1-k-1)d = 2a_1 + (2n-1) d = = a_1 + a_1 + (2n-1) d = a_1 + a_{2n}$

so $a_n + a_{n+1}d = a_1 + a_{2n} = z$

so $a_k + a_{2n+1-k} = z$

so 

 $d \sum_{k=1}^{2n} (a_{k}) $

  $= d \sum_{k=1}^{n} (a_{k} + a_{2n+1-k})$

  $= d \sum_{k=1}^{n} z$

  = 2dnz

 

  So $2dnz = y-x$

    

2021/095) Let $a,\,b,\,c,\,d,\,e,\,f$ be real numbers such that the polynomial

 $P(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f$ factorizes into eight linear factors $x-x_i$ with $x_i>0$ for $i=1,\,2,\,\cdots,\,8$.

Determine all possible values of $f$.

Solution 

Using Vieta's formula  we have

$\sum_{i=1}^8 x_i = 4\dots(1)$

$\sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j = 7\cdots(2)$

$\prod_{i=1}^8 x_i = f\cdots(3)$

We have

$\sum_{i=1}^8 x_i^2= (\sum_{i=1}^8 x_i)^2 - 2 \sum_{i=1}^7 \sum_{j=i+1}^8   x_ix_j $

$= 4^2 - 2 * 7 = 2$

or $\sum_{i=1}^8 x_i^2 = 2$

Subtracting (1) from above

$\sum_{i=1}^8 (x_i^2 - x_i)  = -2$

adding $\frac{1}{4}$ to each term on LHS that is 2 and adding 2 on RHS we get

$\sum_{i=1}^8 (x_i^2 - x_i + \frac{1}{4})  = 0$

or $\sum_{i=1}^8 (x_i - \frac{1}{2})^2  = 0$

so $x_i = \frac{1}{2}$ for each i.

this satisfies the criteria that $x_i$ is positive

putting this in (3) we get $f = \frac{1}{\sqrt[8]2}$