2025/005) Consider the formation of integer, $x = 1555555\cdots 526$ . How to show that whatever amount of digit 5 is placed between $1 \& 2 , 7 ∣ x$

We have

$x = 1555555\cdots 526$ number of 5's >=0$

so $x + 29 = 1555555\cdots 5$

Let it be a n digit number 

so $x + 29 = 10^{n-1} +  5 *111\cdots 1$                                  n-1 1's

or  $x + 29 = 10^{n-1} + \frac{5}{9}(10^{n-1} -1 )$

as 7 is not a factor of 9 

So we have $9(x+29) = 9 * 10^{n-1} + 5 * 10^{n-1} -5$

or $9x + 9 * 29 -5 = 14 * 10^{n-1}$

so $9x + 266$ is divisible by 7

as $266$ is divisible by $7$ so $9x$ is divisible by $7$ and $7$ is coprime to $9$ so $x$ is divisible by $7$

 

2025/004) prove that $\tan \, 1^{\circ}$ is irrational

 We know $\tan\,2A = \frac{2\tan\, A}{1+\tan ^2 A}\cdots(1)$

So if $\tan\, A $ is rational then $\tan 2A$ is rational 

Further $\tan(A+B) = \frac{\tan \, A + \tan\, B}{1-\tan\,A \tan \,B}\cdots(2)$

So if $\tan\, A$ and $\tan \, B$ are rational then $\tan(A+B)$ is rational

So if $\tan\, A$ and $\tan \,nA$ are rational then $\tan(n+1)A$ is rational

using Above 2 we get if $\tan\, A $ is rational then $\tan nA$ is rational for all $n \in Z$

if $\tan \, 1^{\circ}$ is rational then $\tan \, 30^{\circ}$ or $\frac{1}{\sqrt{3}}$ is rational which is not 

Hence $\tan \, 1^{\circ}$ is irrational 

Proved 

 

 

2025/003) If $\frac{1}{a + b}$, $\frac{1}{b + c}$ and $\frac{1}{c + a}$ are in A.P., prove that $b^2, a^2$ and $c^2$ are in A.P.

We have from the properties of AM

$\frac{1}{b + c} - \frac{1}{a + b} =   \frac{1}{c + a} - \frac{1}{b + c}$

Or $\frac{a-c}{(b+c)(a+b)} = \frac{b-a}{((a+c)(b+c)}$

Or $\frac{a-c}{a+b }= \frac{b-a}{a+c}$

Or $(a-c)(a+c) = (a+b)(b-a)$

Or $a^2-c^2 = b^2-a^2$ 

Or $b^2+c^2= 2a^2$   hence $b^2, a^2$ and $c^2$ are in A.P.

2025/002) What is the general solution of the linear diophantine equation $63x-23y=-7$

 Because $GCD(63,23) =1 $ so this has a solution. 

Now let us find a particular solution a and b such that $63a + 23b= 1$

For this we shall use extended Euclidean algorithm.

We have

$63= 2 * 23 + 17\cdots(1)$

$23= 17  + 6\cdots(2)$

$17= 2 * 6 + 5\cdots(3)$

$6= 5 + 1\cdots(4)$

as we have got 1 as the last remainder we are done,.

Now let us work back wards to see

$1= 6- 5$

$=6- (17 - 2 *6)$ from (3)

$= 3 * 6 - 17$

$= 3 * (23-17)  - 17$ from (2)

$= 3 * 23- 4 * 17$

$= 3 * 23 - 4 * (63-2 * 23)$ from (1)

$=11 * 23 - 4 * 63$

So we have $ 1= 11 * 23 - 4 * 63$

Hence $ -7 = - 77 * 23 + 28 * 63$

The above solution is correct however we can reduce the absolute value of coefficients of 23 and 63 by adding 63 to 1st one and subtracting 23 from 2nd one to get

$-7 = 5 * 63- 14 * 23$

so $(x,y) = (5,14)$ is a solution and general solution is $5 + 23t,14+63t)$ as $23 * 63 - 63 * 23 = 0$ 

2025/001) If $GCD(a,35) = 1$ then show that $a^{12} \equiv 1 \pmod {35}$

 We have $35 = 7 * 5$ so

$GCD(a,5) = 1$ and $GCD(a,7) = 1$

Because GCD(a,5) is 1 so as per Fermats Little Thorem $a^{4} \equiv 1 \pmod {5}$ or 

$a^{12} \equiv 1 \pmod {5}\cdots(1) $

Because GCD(a,7) is 1 so as per  Fermats Little Thorem $a^{6} \equiv 1 \pmod {7}$ or

 $a^{12} \equiv 1 \pmod {7}\cdots(2)$

From (1) and 2

 $a^{12} \equiv 1 \pmod {35}$