2020/023) The lowest common multiple of 5 positive integers is 194040. Find the minimum possible sum of these 5 numbers.

The prime factorization of 194040 is:

$194040 = 2^3 ⨯ 3^2 ⨯ 5 ⨯ 7^2 ⨯ 11$

There are 5 numbers which are co-prime to one another.

For the sum of 5 numbers to be lowest the 5 numbers have to be co-prime, The rationale is that each prime factor should have the highest power in at least one number. Say it is x ( x is one of 2,3,5,7,11) Now if it has got any other factor the number becomes bigger and hence the sum

So the numbers are $2^3=8,3^2=9,5, 7^2= 49,11$ and sum is $8+9+5+49+11=82$

2020/022) Find integers $x,y$ such that $x^3-y^3=91$

We have $x^-y-y^3 = 91$

or $(x-y) (x^2+xy+y^2) = 91= 13 * 7$

clearly $x >=y$

So factor of 91 =  1 * 91 and 7 * 13

 so we have following 4 cases

case 1:

$x-y= 1\cdots(1)$

and $x^2 + xy + y^2 =  = 91\cdots(2)$

From (1) we have $x=y+1$

putting in (2) we get $(y+1)^2 + y(y+1) + y^2 = 91$

Or $3y^2 + 3y = 90$

or $y^2+y-30=0$

or $y^2+y-30=0$

or$(y-5)(y+6)=0$

so y = 5 or -6 and x = y+ 1 gives 2 solutions

$(6,5)$ and $(-5,-6)$

case 2:

$x-y= 7\cdots(1)$

and $x^2 + xy + y^2 =  = 13\cdots(2)$

From (1) we have $x=y+7$

putting in (2) we get $(y+7)^2 + y(y+7) + y^2 = 13$

Or $3y^2 + 21y + 49 = 13$

or $3y^2+21y+36=0$

or $y^2+7y+12 =0$

or$(y+3)(y+4)=0$

so y = -3 or -4  and x = y+ 7 gives 2 solutions

$(4,-3)$ and $(3,-4)$

 

case 3:

$x-y= 13\cdots(1)$

and $x^2 + xy + y^2 =  = 7\cdots(2)$

From (1) we have $x=y+13$

putting in (2) we get $(y+13)^2 + y(y+13) + y^2 = 7$

Or $3y^2 + 39y + 169 = 7$

or $3y^2+39y+162=0$

or $y^2+13y+54 =0$

This does not have integer solution

case 4:

$x-y= 91 \cdots(1)$

and $x^2 + xy + y^2 =  = 1\cdots(2)$

From (1) we have $x=y+91$

putting in (2) we get $(y+91)^2 + y(y+91) + y^2 = 1$

Or $3y^2 + 273y + 8281 = 1$

or $3y^2+ 273y+ 8280=0$

or $y^2+91y+ 2760 =0$

ad $91^2-8 * 2760 < 0$ this does not have any real solution

This does not have integer solution

So solution sets are  $(6,5),(-5,-6),(4,-3),(3,-4)$

Short cut solution:(can be applied for objective question)

we need to find the limit of x and y

let is take the difference of $(x+1)^3 and $x^3$ this keeps on increasing when x increases for positive x

and when decreases for -ve x .

so we can bound x between -5 and 6 as $7^3-6^3 > 91$

by putting the value of x from -5 to 6 we can find the value of y as well and solution pair as above 

2020/021) Evaluate $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}$

Let $\theta = \frac{\pi}{7}$

We need to find $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta$ 
Now $7\theta = \pi$

Or $4\theta = \pi - 3\theta$

Taking $\tan$ of both sides we get

Or $\tan 4\theta = \tan (\pi - 3\theta) = - \tan 3\theta$

Or $\frac{4\tan \theta - 4\tan^3 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} - \frac{3\tan \theta - \tan ^3\theta}{1-3\tan ^2 \theta}$

Or  $\frac{4 - 4\tan^2 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} + \frac{3 - \tan ^2\theta}{1-3\tan ^2 \theta}$

 Or  $(4 - 4\tan^2 \theta)(1-3\tan ^2 \theta) + (1-6\tan ^2 \theta + \tan ^4 \theta)(3 - \tan ^2\theta)$

Or $4- 16 \tan ^2 \theta + 12 \tan ^4 \theta + 3  -19 \tan ^2 \theta + 9\tan^4 \theta - \tan ^26\theta = 0$

Or $ \tan ^6\theta  - 21\tan ^4 \theta + 35 \tan ^2 \theta -7=0$

The above equation is a cubic equation in $\tan^2 \theta$ whose roots are $\tan \theta$, $\tan 2\theta$, and 

$\tan 3\theta$,

so using Vieta's formula $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta= 21$ 

or $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}= 21$ 

as a corollary adding 1 to each term of LHS and so adding 3 to RHS we get

 $\sec^2 \frac{\pi}{7}+ \sec^2 \frac{2\pi}{7} + \sec^2 \frac{3\pi}{7}= 24$

 

2020/020) Given a,b,c are positive and $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$ show that $(abc) <= \frac{1}{8}$

we have $\frac{1}{1+a } + \frac{1}{1+b} + \frac{1}{1+c} = 2$
Hence
$(1+b)(1+c) + (1+c)(1+a) + (1+a)(1+b) = 2(1+a)(1+b)(1+c)$
or $1 + b + c + bc + c + a + ac + 1+ a + b + ab = 2 + 2a + 2b + 2c + 2ab + 2bc + 2ac + 2abc$
or $2abc + ab + bc + ca = 1$
Applying AM GM inequality we get
$\frac{1}{4}(ab+bc+ca+2abc) >= \sqrt[4]{2abc.ab.bc.ca}$
Or $\frac{1}{4} >= \sqrt[4] {2 a^3b^3c^3}$
Or $\frac{1}{256} >= 2(abc)^3$
or $(abc) <= \sqrt[3]\frac{1}{512}$
or  $(abc) <= \frac{1}{8}$