2022/055) 2020 distict numbers are in a circle. Show that we can always choose 4 consecutive numbers such that sum of outer 2 numbers is greater than tsom of 2 inner numbers

The 2020 numbers are in a circle. Let the numbers be $a_1$ to $a_{2020}$ now startting wth $a_k$ we have 4 number consecutive $a_k,a_{k+1} ,  a_{k+2}$, $a_{k+3}$ for k from 1 to 2017 and other  4 consecutive  number $a_{2018},a_{2019}, a_{2020}.a_1$ so on

now let us compute the value sum of outer numbers - sum of innter numbers calling $S_n$

$S_{1} = a_{1} - a_{2} - a_{3} + a_{4}$

$S_{2} = a_{2} - a_{3} - a_{4} + a_{5}$

...

...

$S_{2020} = a_{2000} - a_{1} - a{2} + a_3$

If we add all of these then we have

$\sum_{n=1}^{2020}S_{n} = 0$

For the sum of 2020 numbers be zero either all are zero (that means some numbers are same which is not true ) or at least one number is positive say $S_{k}$ then $a_{k}$ is the starting point

 

2022/054) Number $3^{32}-1$ has exactly two divisors greater than 75 and less than 85. What is their product?

we have $3^{32}-1 =  (3^{16}+ 1)(3^{16}-1) =   (3^{16}+ 1)(3^{8}+1) (3^{8}- 1)$

$ = (3^{16}+1)(3^8+1)(3^4+1)(3^4-1)$ 

out of the above $3^4+1 = 82$ and %3^4-1=80$ are between 75 and 85 and product is 80 * 82 = 6560

2022/053) If $x^2-x=1$ Symplify $\frac{x^5+8}{x+1}$

 We have $x^2 = 1-x\cdots(1)$

multiply both by x to get

$x^3 = x -x^2 = x - (1-x) = 2x -1\cdots(2)$

 $x^5 = x^3 .x x^2 = (2x-1)(1-x) = -2x^2 +3x -1 = -2(1-x) + 3x - 1 = 5x-3$ (using (1))

or $x^5 + 8 = 5x + 5$

or  $\frac{x^5+8}{x+1} = 5$

2022/052) Simplify expression $(\frac{1}{2^3-1} + \frac{1}{2}) (\frac{1}{3^3-1} + \frac{1}{2}) \cdots (\frac{1}{100^3-1} + \frac{1}{2})$

We have let $t_n =   \frac{1}{t^3-1} + \frac{1}{2}$

$= \frac{1}{2}\frac{t^3+1}{t^3-1}$

=$\frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$

Now given value = $\prod_{n=2}^{100} \frac{1}{2}\frac{(t+1)(t^2-t+1)}{(t-1)(t^2+t+1)}$

$= \frac{1}{2^{99}} \prod_{n=2}^{100} \frac{(t+1)}{(t-1)}\frac{(t^2-t+1)}{(t^2+t+1)}$

now let us compute the 2 products 

$ \prod_{n=2}^{100} \frac{(t+1)}{(t-1)}= \frac{3}{1}.\frac{4}{2}.\frac{5}{3}\cdots \frac{98}{96}. \frac{99}{97}. \frac{100}{98}. \frac{101}{99}$

In the above product the numerator of $n^{th}$ term is same as denoniator of $n+2^{nd}$ term giving

$ \prod_{n=2}^{100} \frac{(t+1)}{(t-1)} = \frac{100 * 101}{1 * 2}\cdots(1)$

let us compute 2nd product

$  \prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{7} ( \frac{ 7}{13} \cdots \frac{9901}{10101}$

In the above product the numerator of $n^{th}$ term is same as denoniator of $n-1^{st}$ term giving

$  \prod_{n=2}^{100} \frac{(t^2-t+1)}{(t^2+t+1)}= \frac{3}{10101}\cdots(2)$

using (1) and (2) we get the required result

$= \frac{1}{2^{99}}\frac{3}{10101} * 5050= \frac{2525}{2^{98} * 3367}$

2022/051) Show tthat there are infnite solutions to $x^3+y^4= z^{31}$

We know that 2^m * 2 is a power of 2

so if we chose $x^3 = y^4 = 2^t$ for some t then we shall get power of 2

so let $x = 2^{4m}$ and $y = 2^{3m}$

so $x^3 + y^ 4= 2^{12m} + 2^{12m} = 2^{12m+1}$

 Now there is only one 1ssure remanins that is 12m + 1 has to be multiple of 31

that is not a problem as gcd(31,12) is 1 and we shall use exteneded euclidean algoirithm to find the same

now 31 =  2 * 12 + 7 we take the highest multiple of 12 and leave the remainder

or $7 = 31 - 2 * 12\cdots(1)$

12 = 7 * 1 + 5

or $5 = 12 - 7 * 1 = 12 - (31 - 2 * 12)$ using (1)

or  $5 = 12 * 3 - 31\cdots(2)$

now $ 7 = 5 + 2$

or $2 =  7- 5 = (31- 2 * 12) - (12 * 3 -31) = 2 * 31 - 5 * 12$ (uinsg (1) and (2)

or $2 = 2 * 31 - 5 * 12$

now $5 = 2 * 2 + 1$

or $1= 5 - 2 * 2$

or $1 = (12 * 3 - 31) - 2 * ( 2 *31 - 5 *12) = 13 * 12 - 5 * 31$

so if m is 13 or any number 31n + 13 we get 12m+ 1 is a multiple of 31

m = 31n + 13 gives 12m +1 = 12(31n + 13) = 31(12n + 5)

so $x = 2^{4(31n + 13)}$, $y = 2^{3(31n + 13)}$ and $z = 2^{12n+5}$ is parametric solution and hence infinite solutions.

2022/050) Find n such that $n! = 3!5!7!$

 clearly n > 7.

now if t ia factorial then 3!5! must be product of sum k sucesssive numbers from 8

3!5! = 6 * 120 = 720 = 8 * 90 = 8 * 9 * 10

so n! = 10 * 9 * 8 * 7! = 10! 

or n = 10

2022/049) Prove that if a number is sum of 3 squares then square of the number is sum of 3 squares

 Let n be sum of 3 squares so

$n= a^2 + b ^2 + c^2$

we need to show that $n^2$ is sum of 3 squares

from given condition

$n^2 = (a^2+b^2+c^2)^2 = a^4 + b^4 + c^4 + 2 a^2b^2 +2 b^2 c^2 + 2 c^2 a^2$

$=  a^4 + b^4 + c^4 + 2 a^2b^2 -2 b^2 c^2 - 2 c^2 a^2 + 4 b^2c^2 + 4 c^2 a^2$

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

sum of 3 sqaures and hence proved 

 

2022/048) Given $\frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1$ Find $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

 We are given

$\frac{a^2}{b+c} = 1$

or $a^2 = b + c$

or $a^2 + a = a + b + c$

or $a(a+1) = a + b + c$

or $\frac{1}{a+1} = \frac{ a}{a + b+ c}\cdots(1)$

similarly  $\frac{1}{b+1} = \frac{a}{a + b+ c}\cdots(2)$

and  $\frac{1}{c+1} = \frac{a}{a + b+ c}\cdots(3)$

adding all 3 above we get  $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{a+b+c}{a+b+c} = 1$