2024/008) Find minimum Value of $(a+7)^2+(b+2)^2$ with Constraint $(a-5)^2+(b-7)^2=4$

We have the constraint  $(a-5)^2+(b-7)^2=4$ this is set of points lying in a circle with centre (5,7) and radius 2.

We need to find the minimum of  square of the distance of (a,b) from (-7,-2) and this is minimum when  (a,b) lies in the line from (5,7) to (-7,-2)

Distance from (5,7) to (-7,2) = $\sqrt{(5+7)^2 + (7+2)^2}= \sqrt{12^2 + 9^2} = 15$

So distance from (a,b) to (-7,  -2) is $15-2 = 13$

Minimum Value of $(a+7)^2+(b+2)^2= 13^2 =169$

2024/007) Find the value of k such that $(\frac{1}{x+k} + \frac{k}{x-k} + \frac{2k}{k^2-x^2})(| x-k | -k) = 0$ has exactly one non -ve root

Reference: https://www.physicsforums.com/threads/discovering-the-solution-problem-of-the-week-261-apr-23rd-2017.1037211/

Note that we cannot have or . Multiply both sides by , we get or As cannot be we have , this gives 2 values of , where or . Hence there is no solution to the problem.

Reference: https://www.physicsforums.com/threads/discovering-the-solution-problem-of-the-week-261-apr-23rd-2017.1037211/

Note that we cannot have $x=k$ or $x = -k$ 

So multiplying both sides by $x^2-k^2$

We get 

$((x-k) + k(x+k) + 2k) )(| x-k |  -k) = 0$

Or $(x+k)(k+1))(| x-k |  -k)=0$

As x cannot be -k  $(| x-k |  -k)=0$

So we get 2 values of x that is 2k or 0

Hence no solution

2024/006) Show that $10^{th}$ digit of a power of 3 is always even

The power of digit of 3 shall have 1 or 3 or 7 or 9 as the unit digit.

Let us see some power of 3 that is 1,3,27,81(after this the sequence in the 1st digit repeats.

So we have (20n + x) where x is 1 or 3 or 7 or 9.

When we multiply by 3 we get 60n + 3 or 60n + 9 or 60n + 21 or 60n + 27 that is 60n + 3 or 60n + 9 or 20(3n+1) + 7 or 20(3n+1) + 9

So tens digit is even

 

    

2024/005) For m > 1 show that if $2^{2m+1} > n^2$ then $2^{2m+1} \ge n^2+7$

Because $m > 1$ $2m + 1\ge 4$  So    $2^{2m+1}$ is divisible by 16.

  $2^{2m+1} $ is double of $(2^m)^2$ and is not a square.

Now we consider 2 cases

Let us consider when n is odd 

Then $n= (2k+1)$

So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1)+1$

Ss k or k+1 is even we $n^2 \equiv 1 \pmod 8$ 

So $2^{2m+1}$ is divisible by 8 and minimum number above $n^2$ which is divisible by 8 is $n^2 + 7$

So we have proved for the case n is odd.

Let us consider when n is even

If  $n^2$ is divisible by 8 then $n^2$ is divisible by 16 and as $2^{2m+1}$ is divisible by 16 and n^2 is divisible by 16 

So $2^{2m+1}-n^2 = 16k$ for some positive k

So $2^{2m+1}-n^2 \ge16$

Hence  $2^{2m+1}-n^2 \gt 7$ or $2^{2m+1} \gt n^2+ 7$

If $n^2$ is even and is not divisible by 8 then n is of the form 4k+2

$n^2= 16k^2 + 16k + 4$

Or $n^2 \equiv 4 \pmod {16}$ 

 As $2^{2m+1} = 0 \pmod {16}$

So $2^{2m+1} \ge n^2 + 12 $

Hence  $2^{2m+1} \gt n^2 + 7 $

 we have proved for all 3 cases hence done

 

 

  

2024/004) Find the smallest a and b such that $\frac{1}{640} = \frac{a}{10^b}$

We have $640= 128 * 5 =  2^7 * 5$

To make it a power of 10 we need to multiply by $5^6$ giving $10^7$

so $a=5^6$ and b = 7 are the smallest values

2024/003) Find $\sum_{n=1}^{\infty} \frac{n^2}{10^n}$

Here let $frac{1}{10} = x$

 We have for $|x|  \lt 1$

 $\sum_{n=0}^{\infty} x^n= \frac{1}{1-x}$

Differentiating both sides wrt x we get

$\sum_{n=0}^{\infty} nx^{n-1}= \frac{1}{(1-x)^2}$

Multiplying both sides by x we get 

 $\sum_{n=0}^{\infty} nx^n= \frac{x}{(1-x)^2}$

Differenting both sides wrt x we get

$\sum_{n=0}^{\infty} n^2x^{n-1}= \frac{d}{dx} \frac{x}{(1-x)^2}$

Usimg $\frac{d} ({dx} \frac{u}{v})= \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

We get 

 $\sum_{n=0}^{\infty} n^2x^{n-10} =\frac{(1-x)^2 . 1 - x . (x-2)}{(1-x)^4} = \frac{x +1}{(1-x)^3}$

Multiply by x on both sides we get $\sum_{n=0}^{\infty} n^2x^n = \frac{x^2 +x}{(1-x)^3}$

Putting value of $x = \frac{1}{10}$ we get the result

2024/002) If x = $2^{\frac{1}{3}} + 2^{-\frac{1}{3}}$ , How do you prove that $2x^3-6x = 5$

 Let $a =  2^{\frac{1}{3}}\cdots(1)$

So $a^3 = 2\cdots(2)$ 

And $\frac{1}{a} = 2^{-\frac{1}{3}}\cdots(3)$

And  

$x = a+ \frac{1}{a}\cdots(4)$

Cubing both sides $x^3 = a^3 + \frac{1}{a^3} + 3 * a * \frac{1}{a} ( a + \frac{1}{a})$

Or $x^3 = a^3 + \frac{1}{a^3} + 3 * ( a + \frac{1}{a})$

Or  $x^3 = 2 + \frac{1}{2} + 3 *x$ from (2), 3) and (4)

Or $x^3 =  \frac{5}{2} + 3 *x$

Or $2x^3 = 5 + 6x$

Or $2x^3-6x = 5$

Proved

2024/001) Given $\frac{x-a}{x-b} + \frac{x-b}{x-a} = \frac{a}{b} + \frac{b}{a}$

Let  $\frac{x-a}{x-b} = m$ and $\frac{a}{b} =n $ 

So we get   $ m + \frac{1}{m} = n + \frac{1}{n}$

Or $m^2n + n = n^2m + m$

Or $m^2n - n^2m = m - n$

Or $mn(m -n ) = m - n$

Or  m-n = 0 Or $mn = 1$

case 1

m = n gives $\frac{x-a}{x-b} = \frac{b}{a}$

Or $b(x-a) = x(a-b)$ or $bx-ab = ax-ab$ or $bx = ax$ or $x(a-b)=0$ or $x = 0$ 

Case 2)

mn =1 gives 

$\frac{x-a}{x-b}*\frac{a}{b}=1$

or $a(x-a)=b(x-b)$

Or $ax-a^2=bx-b^2$

Or $ax-bx  = a^2-b^2$

Or $x(a-b) = a2-b^2$

Or $x = a + b$