2020/029) Solve in positive integers $ab - 4a -4b = -8$

Because it is symmetric in a ,b without loss of generality we can take $a>b$ and find a solution for the same
then permutation shall be another solution.

We need to factor the same in some form and as it is of the form $ab-4a-4b$ this shall have the form $(a-4)(b-4)$
Now $(a-4)(b-4) = ab -4a -4b + 16$
Or $(a-4)(b-4) = -8 + 16$ Putting the value of ab-4a-4b from given condition
Or $(a-4)(b-4) = 8$
For a and b to be integers we need to take the factors of 8 giving 2 sets 8 = 8 * 1 or 4 * 2
$a-4=8,b=4=>a=12, b= 5$

and
$a-4=4,b=2=>a=8, b= 6$

 

2020/028) Compare $\sqrt[n+1]{(n+1)!}$ and $\sqrt[n]{(n)!}$ where n is integer

 We have $(n+1)!= (n+1)n!$

Taking $n^{th}$ power we have $((n+1)!)^n = ((n+1)n!)^n = (n+1)^n (n!)^n\cdots(1)$
Now $(n+1)^n > n!\cdots(2)$
From (1) and (2) we get $((n+1)!)^n > (n!)^{n+1}$
Taking $(n(n+1))^{th}$ root we get $\sqrt[n+1]{(n+1)!} > \sqrt[n]{(n)!}$

2020/027) Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer n.

1st we see that $n^2 + 5n + 23 = n(n+5) + 23$ which is odd so

it is not divisible by 2

We have $4n^2 + 20n + 92 = (2n + 5)^2 + 67$
So we have for x = 1 we get 67 + 1 = 68 = 4 * 17
So we need to check for odd numbers upto 17 and can see if we have any smaller factor less that 17. this is based on the fact that if it divisible by p then we can always have a number $n < p$ for which it is true.
x = 3 67 + 9 = 76 = 4 * 19
x=5 67 + 25 = 92 = 4 * 23
x =7 67 + 49 = 116 = 4 * 29
x = 9 67 + 81 = 148 = 4 * 37
x =11 67 + 121 = 188 = 4 * 47
x = 13 67 + 169 = 236 = 4 * 59
x = 15 67 + 225 = 292 = 4 * 73
from the above we find that 17 is the smallest number for which we have
x = 1
or $2n + 5 \equiv 1 \pmod {17}$
solving this n = 15 and for n = 15 we have

If it divides $n^2 + 5n + 23$ then if divides $4n^2 + 20n + 92$ and no smaller prime can divide the same as the number is coprime to 4

Now let us take square of odd number add to 67 that is $x^2+67$

So the smallest prime is 17

$n^2 + 5n + 23 = 15^2 + 5 * 15 + 23 = 323 = 17 * 19$    

2020/026) Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric number.

 Let the common ratio of the digits of smallest number be x.

The smallest digit is 1 and largest digit is 9 and the ratio of them is 9 which is $3^2$

So we have a number 139 starting with 1 and let us check if can have a smaller number than this. if take the ratio 2 then we get a smaller number 124.
Now that larger number is 931 with the common ratio with $\frac{1}{3}$ and as we do not have a factor of 9 more than 3( excluding 9 it self) the largest number is 931.
So the difference is $931-124= 807$

2020/025) How many positive perfect squares less than $10^6$ are multiples of 24?

 Because the number is a perfect square and multiple of 24 this should be multiple of smallest multiple of 24 which is a square.

Now  $24 = 2^3 * 3 $ so to get the smallest multiple of 24 which is a square we need to make the power of 2 and 3 both even or we need to multiply by 2* 3 or 6 to get 144. So we need

$144n^2 = (12n)^2 <= 10^6$ 

or $12n <= 1000$

or $n <= \frac{1000}{12}= 83.3$

So $n < =83$ and hence there are 83 numbers

2020/024) Show that if for sides a,b,c of a triangle if for each integer n the sides $a^n,b^n,c^n$ form a triangle then the triangle is isosceles.

Without loss of generality let us assume that a is the longest side. if the triangle is not  isosceles then we have $ a > b$ and $a > c$.

For it to an triangle we need to have $a^n < b^n + c^n$

Or $$(\frac{b}{a})^n +  (\frac{c}{a})^n > 1\cdots(1)$$ for all n

As $b < a$ so $(\frac{b}{a}) < 1$ and  $(\frac{b}{a})^n < 1$ and as n goes to infinity this goes to zero. 

Similarly $(\frac{c}{a})^n < 1$ and as n goes to infinity this goes to zero. 

So the sum goes to zero and hence (1) is not true so $a^n,b^n.c^n$ sides cannot form a triangle

So  either b or c has to be same as a. So the triangle is isosceles.

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  $(abc) <= \frac{1}{8}$

Q2020/019) If $3^x=4^y=12^z$, then prove that $\frac{1}{x} + \frac{1}{y}= \frac{1}{z}$

$3^x = 12^ z$

so $3 = 12^\frac{z}{x}\cdots(1)$

$4^y = 12^z$

So $4^y = 12^ \frac{z}{y}\cdots(2)$

Hence $12 = 3 * 4 = 12^{(\frac{z}{x} + \frac{z}{y})}$ From (1) and (2)

So $\frac{z}{x} + \frac{z}{y}= 1$ or $\frac{1}{x} + \frac{1}{y}= \frac{1}{z}$

2020/018) Prove that $3^n>=27n^3 $ for $n >=9$

To prove the same we use principle of mathematical induction

Base step

For n = 9 $LHS = 3^9 = 3^3 * 3^6 = 27 * 9^3$ so $3^n >= 27n^3$

SO base step is true

Now $(\frac{n+1}{n})^3$ decreases as n increases and at n = 9 we have   $(\frac{n+1}{n})^3= \frac{1000}{729}< 3$

So $(\frac{n+1}{n})^3< 3$ for all $n>=9$

Or $3 > (\frac{k+1}{k})^3\cdots(1)$ for all $k>=9$

Let it be true for n = k $k >=9$

We need to prove it to be true for n =  k+ 1

$3^k > = 27 k^3$

Multiplying by (1) on both sides

$3^{k+1} > = 27 * (\frac{k+1}{k})^3 * k^3 $

Or $3^{k+1} >= 27(k+1)^3$

So it is true for n = k+ 1

We have proved the induction step also

Hence proved

2020/017) FInd q when the equation $x^4-40x^2+q=0$ has 4 roots is AP.

Because it has 4 roots in AP so let the roots be $a-3d, a-d,a+d, a+3d$ 
The sum of the root is zero as coefficient of $x^3=0$
So we have $(a-3d)+(a-d)+(a+d)+(a-3d) = 4a = 0$
Or $a=0$
Hence the roots are $-3d, -d, d, 3d$
So Equation becomes  $(x+3d)(x+d)(x-d)(x-3d)=0$
Or $(x+3d)(x-3d)(x+d)(x-d)=0$
Or $(x^2-9d^2)(x^2-d^2) =  x^4-10d^2x^2+ 9d^4=0$
Comparing with given equation $-10d^2= - 40$ or $d^2=4$
And $q=9d^4= 9 (4)^2= 144$
Hence $q= 144$

2020/016) find n such that $\sqrt{n} + \sqrt{n+2005}$ is a natural number

Let $\sqrt{n+2005} + \sqrt{n} =m \cdots(1)$

We know $(n+2005) - (n) = 2005$
Or $(\sqrt{n+2005})^2 - (\sqrt{n})^2 = 2005$
Or $(\sqrt{n+2005} +  \sqrt{n}) (\sqrt{n+2005} - \sqrt{n}) = 2005\cdots(2)$

Dividing (2) by (1) we get
$(\sqrt{n+2005} - \sqrt{n}) = \frac{2005}{m}\cdots(3)$

Subtracting  (3) from (1) we get $2\sqrt{n}= m - \frac{2005}{m}$

Clearly we have $m^2>=2005$ so we choose m factor of 2005 that is 401, 2005

Taking $m= 2005$ we get $n = (\frac{1}{2}(2005-1)^2 = (1002)^2 = 1004004$

Taking $m= 401$ we get $n = (\frac{1}{2}(401-5)^2 = (198)^2 = 39204$

So we have solution set $(1004004,39204)$

2020/015) Find integers n and k such that $n!+8 = 2^k$

Firstly $n\lt 6$ because if $n >= 6$ then is is divisible by $2^4$ ( 2 comes once in 2 twice in 4 and once is $6$ so at least 4 times and $2^4=16$.

we can show it in another way that $6!=720 = 16 * 45$
so $n>6$ $n!$ shall be divisible by 16 so say value is 16m

now $n!+8 = 16m+8 = 8(2m+1)$ and as it is product of 8 and an odd number greater than 1 so it cannot be power or 2 .

Further n cannot be less than 4 because 8 has to  to factor of n

so we need to check for n = 4 and n = 5

no $4!+8 = 24 + 8 = 32$ giving n =  4 k = 5

$5! + 8 = 120 + 8 = 128 = 2^7$ giving n =  5 and k = 7

Q2020/14) Solve in natural numbers $\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}$

we get
$2018b + 2018 a = 3ab$
or $3ab - 2018 a - 2018b = 0$
As the coefficient of a,b are same so we get product of 2 monomials one of a one of b and coefficients being same.
Multiplying by 3 we get
$9ab - 3 * 2018 a - 3 * 2018b = 0$
or $9ab - 3 * 2018 a - 3 * 2018b = 2018 ^2 $
or $(3a-2018)(3b-2018) = 2018^2= 2^2 * 1009^2$
LHS each term is 1 mod 3 so we should make the RHS so product of 2 numbers each of which is 1 mod 3.


if(a,b) is a solution then (b,a) is also a solution so let us assume that $a>=b$

so we get following sets

$3a-2018= 2018^2, 3b-2018=1$  giving $(a= 1358114,b= 673)$
$3a-2018= 1009^2 , 3b=2018=4$ giving $(a= 340033, b= 674)$
$3a-2018= 1009, 3b-2018=1009 * 4$ giving $(a= 1009,b= 2018)$

and permutation of the same.

2020/13) Solve $2\log_2(x+15) - \log_2x = 6$

We have
$2\log_2(x+15) - \log_2x = \log_2(x+15)^2 - \log_2(x) = \log_2\frac{(x+15)^2}{x} = 6$
Or $\frac{(x+15)^2}{x} = 2^6= 64$
Or $(x+15)^2 = 64x$
Or $x^2+30x+225-64x=0$
Or $x^2-34x+ 225=0$
Or $(x-25)(x-9)= 0$ So x = 25 or 9 

2020/012) Solve in real $x^4+y^4+z^4 + 1 = xyz$

We have LHS is positive so RHS is also positive.

So all of x,y,z are positive or one is positive and rest 2 are negative

So let us solve for positive x,y,z

Using AM GM inequality we have  $\frac{x^4+y^4+z^4+1}{4} >=\sqrt[4]{x^4y^4z^4}$

Or $x^4+y^4+z^4 +1 >= 4xyz$

And they are equal if $x=y=z$

So we get $3x^4-4x^3+1 = 0$

Or  we have trying $x=1$ and $x = 3$ x=1 is solution

giving one set of solution = $(1,1,1)$ and 2 negative  and one positive gives 3 more $(1,-1,-1), (-1,-1,1), \,and\, (-1,1,-1)$

2020/011) Prove that $4^{79}< 2^{100} + 3^{100} < 4^{80}$

We need to prove 2 inequalities
$4^{79}< 2^{100} + 3^{100} $ and $2^{100} + 3^{100} < 4^{80}$

For

$4^{79}< 2^{100} + 3^{100} $

If we prove that  $4^{79} < 3^{100} $ we are through

We have $\frac{4^{80}}{3^{100}} = (\frac{4^4}{3^5})^{20} = (\frac{256}{243})^{20}\cdots(1)$

Now $\frac{256}{243} = 1+\frac{13}{243} < 1 + \frac{13}{13 *18} = 1 + \frac{1}{18} = \frac{19}{18}\cdots(2)$

Hence $(\frac{256}{243})^{20} < (\frac{19}{18})^{20} = (\frac{361}{324})^{10}=  (1 + \frac{37}{324})^{10}\cdots(2)$

We have
$\frac{37}{324} < \frac{37}{37 * 8} < \frac{1}{8}$

From (2) and above
$(\frac{256}{243})^{20} < (1 + \frac{1}{8})^{10} = (\frac{9}{8})^{10} = (\frac{81}{64})^5 < (\frac{81}{63})^5 = (\frac{9}{7})^5 = \frac{59049}{16807} < 4$

From (1) and above $\frac{4^{80}}{3^{100}} < 4 $ and hence $4^{79} < 3^{100}$

Hence $4^{79} < + 2^{100} + 3^{100}$

For proving the $2^{nd}$ part we have

  $(\frac{256}{243})^20 = (1+  \frac{13}{243})^{20} > 1+  \frac{13}{243} * 20  > 1+ \frac{260}{243} > 2$ by bionomial expansion and deleting positive terms

Form above and (1) we have

$\frac{4^{80}}{3^{100}} > 2$ or $4^{80} > 2 * 3^{100} > 3^{100} + 2^{100} $

Proved

2020/010) If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero.

The $\frac{a}{b}$ equals  a. $\frac{n-5}{6}$ b. $\frac{n-4}{5}$ c. $\frac{5}{n-4}$ d. $\frac{6}{n-5}$ 

Solution:
we have

$5^{th}$ term = ${n \choose 4} a^{n-4} (-b)^4$
$6^{th}$ term = ${n \choose 5} a^{n-5} (-b)^5$

As sum of the two is zero we have ${n \choose 4} a^{n-4} (-b)^4 + {n \choose 5} a^{n-5} (-b)^5 = 0$

or ${n \choose 4} a - {n \choose 5} (b) = 0$

or $\frac{a}{b} = \frac{n \choose 5}{n \choose 4}$

$= \frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}}$

$= \frac{(n-4)!}{(n-5)! 5}= \frac{n-4}{5}$

Hence ans (b)

2020/009) Solve in real $(x+4)(x+5)(x+6)(x+7) = 1680$

We have as  4+7 = 5 + 6

$((x+4)(x+7))((x+5)(x+6)) = 1680$
Or $x^2+11x+28)(x^2+11x+30) = 1680$
Putting $x^2+11x + 29= t$ we get
$(t-1)(t+1) = 1680$
Or $t^2 -1 = 1680$ or $t^2 = 1681$ or $t=\pm 41$

So we get 2 equations $x^2+11x+29=41$ or $x^2+11x+29=-41$

1st equation gives $x^2+11x-12=0$ or $(x+12)(x-1) = 0$ giving $x = -12,1$

$x^2+11x+29=-41$ giving $x^2+11x+70=0$ does not have real root

2020/008) A right angled triangle has perimeter 40 m and area 60 $m^2$ Find the lengths of the sides of the triangle.

Let the 3 sides be a,b & c, where c is the hypotenuse and without loss of generality $a>b$.
We have as it is right angled triangle
$a^2+b^2 = c^2\cdots(1)$
As perimeter is 40 we have a+b+c = 40 and hence
$a+b= 40 -c\cdots(2)$
Area is 60 so we have
$\frac{ab}{2} = 60$ or $ab=120\cdots(3)$
From (2) and (3) $(a+b)^2 - 2ab = (40-c)^2 - 240 $
Or $a^2+b^2 = c^2 - 80c + 1600 - 240 = c^2 - 80c + 1360 $
Or $80c-1360 = c^2-(a^2+b^2) = 0$ sing (2)
Or c = 17
Puttng c =17 in (2) we get
$a+b= 40-17 = 23$
So $(a+b)2^ = 23^2$
Using above and (3) we get $(a-b)^2 = (a+b)^2 - 4ab = 23^2 - 4 * 120 = 49$
Or a-b = 7\cdots(4)
From (2) and (4) a = 15 and b = 8
So sides of triangle are 15m,8m,17m


This question I picked from https://in.answers.yahoo.com/question/index?qid=20200310093657AAFSsOP

2020/007) We know for a right angled triangle $a^2+b^2=c^2$ where a,b are shorter sides and c is hypotenuse. What is $\frac{1}{a^2}+ \frac{1}{b^2}$ for a right angled triangle

We have $a^2+b^2=c^2$

Hence deviding by $a^2b^2$ we get

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

Now area of the triangle $A = \frac{1}{2}ab\cdots(2)$

If h is the altitude drawn from the right angle to the hypotenuse then area of
the triangle $A = \frac{1}{2}ch\cdots(3)$

So from (2) and (3) ab = ch and putting in (1)

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

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

Where h is the altitude drawn from the right angle to the hypotenuse

2020/006) How many squares and rectangles are there on a standard chess board

First let us calculate the number of squares


For the square of side n we can choose in rows in 9-n ways(side 1 8 ways, side 2
7 ways son on). In the column in 9-n ways. so the number of ways the
square of side n can be chosen in $(9-n)^2$ ways and as we have number of sides from 1 to 8
so number of ways = $$\sum_{k=1}^{8}(9-k)^2= \sum_{n=1}^{8}(n)^2 = 204$$

Now for calculation of number of rectangles.


For a rectangle we need to choose 2 lines in rows $9 \choose 2$ ways and in
columns in $9 \choose 2$ ways so total number of ways ${9 \choose 2  }^2$ or
$(\frac{9 * 8}{2})^2$ ways that is 1296 ways. As there are 204 squares
so number of rectangles = 1296-204 = 1092

2020/005) Compare $\log_23$ and $\log_35$

We have $2^3=8 < 9 = 3^2$
Hence $2^\frac{3}{2} < 3$
Or $\log_23 > \frac{3}{2}\cdots(1)$
$3^3=27 > 25 = 5^2$
Hence $3^\frac{3}{2} > 5$
Or $\log_35 <  \frac{3}{2}\cdots(2)$
Using (1) and (2) $\log_23 > \log_35$

2020/004) Prove that $\log_abc.\log_bca.\log_abc = 2 + \log_abc+\log_bca+\log_abc$

Solution
Let $\log_ab= x\cdots(1)$
$\log_bc=y\cdots(2)$
$\log_ca=z\cdots(3)$

So we have $xyz= \log_ab.\log_bc.\log_ca=1 \cdots(4)$
$\log_abc= \log_ab + \log_ac = x + \frac{1}{z}= x + xy= x(1+y)\cdots(5)$
Similarly
$\log_bca= y(1+z)\cdots(6)$
And
$\log_bca= z(1+x)\cdots(7)$


Now $LHS= \log_abc.\log_bca.\log_abc$
$=x(1+y) . y(1+z) . z (1+x)$ from (5),(6),(7)
$= xyz(1+y)(1+z)(1+x)$
$= (1+y)(1+z)(1+x)$ from (4) as xyz=1
$= 1 + y + z + x + xy+yz + zx + xyz$
$= 1 + y + z + x + xy+yz + zx + 1$ from (4) as xyz=1
$= 2 + y + yz + x + xy+ z + zx$ Rearrangement of terms
$=2 + y(1+z) + x(1+y) + z(1+x)$
$=2 +\log_bca + \log_abc + \log_cab$
$=2 + \log_abc + \log_bca + \log_cab$  Rearrangement of terms
$=RHS$


Hence Proved

2020/003) If 1 < x < 2, Then Prove that $\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}} = \frac {2}{2-x}$

Solution

Let $\sqrt{x+2\sqrt{x-1}} = \sqrt{m} + \sqrt{n}$
Hence  $\sqrt{x-2\sqrt{x-1}} = \sqrt{m} - \sqrt{n}$
Without loss of generality let us assume $m >=n$
Square both sides to get  $x+2\sqrt{x-1} = m + n + 2\sqrt{mn}$
Comparing rational and surds on both sides we get
$x=m +n\cdots(1) $
$x-1= mn\cdots(2)$
From (1) and (2)
$(m-n)^2 = (m+n)^2-4mn = x^2-4(x-1) = (2-x)^2$ we chose 2-x as $x < 2$
or $m-n=2-x\cdots(3)$
From (3) and (1) we get $m=1, n = x-1$
Hence we get
$\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}}$
$= \frac{1}{\sqrt{m}+\sqrt{n}}+ \frac{1}{\sqrt{m}-\sqrt{n}}$
$= \frac{\sqrt{m}+\sqrt{n}+ \sqrt{m}-\sqrt{n}} {m-n}$
$= \frac{2\sqrt{m}} {m-n}$
$= \frac{2\sqrt{1}} {2-x}$
$= \frac{2} {2-x}$

2020/002) a,b,c are in AP, x,y,z are in HP and ax,by,cz are in GP then prove that $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$

We have $x,y,z$ are in HP so
$\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$
Squaring both sides
$\frac{1}{x^2} + \frac{1}{z^2} + \frac{2}{xz} = \frac{4}{y^2}$
Or $\frac{1}{x^2} + \frac{1}{z^2} = \frac{4}{y^2} - \frac{2}{xz}  $
Or $\frac{x^2+z^2}{x^2z^2} = \frac{4}{y^2} - \frac{2}{xz}  $
Or $\frac{x^2+z^2}{xz} = \frac{4xz}{y^2} - 2$\
Or $\frac{x}{z} + \frac{z}{x} = \frac{4xz}{y^2} - 2\cdots(1)$
As ax,by,cz are in GP
$axcz = b^2y^2$
Or $\frac{xz}{y^2} = \frac{b^2}{ac}$
Putting above in (1)
$\frac{x}{z} + \frac{z}{x} = \frac{4b^2}{ac} - 2\cdots(2)$
As a,b,c are in AP we have $2b= a + c$
Putting in (2) we get
$\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2}{ac} - 2$
Or $\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2-2 ac} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a^2+c^2} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$
proved

2020/001) a,b,c are in AP, b,c,d are in GP, and c,d,e are in HP, prove that a,c,e are in GP

Solution

We are given a,b,c are in AP so
$2b= a +c \cdots(1)$
b,c,d are in GP so
$bd=c^2\cdots(2)$
c,d,e are in HP, so
$\frac{1}{c} + \frac{1}{e} = \frac{2}{d}$
or $\frac{e+c}{ce} = \frac{2}{d}\cdots(3)$
or $d = \frac{2ce}{e+c}$
From(2)
$c^2 = bd$
or $2c^2 = 2bd = (a+c) \frac{2ce}{e+c}$ putting from (1) and (3)
or $c^2(e+c) = ce(a+c)$
or $c(e+c) = e(a+c)$
or $c^2= ae$

Hence  a,c,e are in GP