2025/008) Find prime numbers x,y,z such that $7(x+y+z) = xyz$

Because $x,y,z$ are prime one of $x,y,z$ must be 7 so without loss of generality we assume $z = 7$

Putting in original equation we have

$7(7+x+y) = 7xy$

Or $xy - x -y -7=0$

Or $xy-x-y +1 = 8$

Or $x(y-1) - (y-1)=8$

Or $(x-1)(y-1)= 8= 1 * 8 = 2 * 4$

We assume $x \ge y$

this gives $x-1=8$ or x= 9 in which case it is not prime so not proper

Or $x-1=4,y=1=2$ giving $x = 5,y=3$ giving $x=5,y=3,z=7$ or any permutation of the same

2025/007) If 2n+1 and 3n+1 are both perfect squares, then how could we prove that 40|n?

For n this to be divisible by 40 we need to show that it is divisible by 8 and 5 as 40 = 8 * 5 and 8 and 5 are co-primes  

 First let us prove that 8 divides n

Clearly 2n+1 is odd so this has to be square of odd number say 2k+1

Now $(2k+1)^2 = 4k^2 + 4k +1=4k(k+1)+1 = 4m+1$ when $m= k(k+1)$

or $2n+1 = 8m+1$ or $n= 4m$

so n is even and hence 3n+1 is odd so  square of odd number say 2p+1

$3n+1 = (2p+1)^2 = 4p^2 + 4p +1 = 4p(p+1) +1$

or $3n = 4p(p+1)$ as p(p+1) is even it can be written as 2q

as $3n+1 = 8q+1$5

or $3n = 8q$

So 8 divides 3n and as 8 and 3 are copimes so 8 divides n 

now let us prove that 5 divides n

let us find what are the numbers that can be remainder of $x^2$ when divided by 5

we have x of the form $ 5q+k$ where k is one of $0,\pm 1,\pm 2$

$x^2= 25q^2 + 10kq +k^2$

$= 5(5q^2 + 2k) + k^2$

so remainder when $x^2$ divided by by5 is 0 or 1 or 4

now let us find n for which 2n+1 give a remainder 0 or 1 or 4 else it cannot be a square.

$n = 5k$ gives $2n + 1 = 10 k  + 1 = 5 * 2k + 1$ remainder is     1 which is a valid candidate

$n = 5k+ 1$ gives  $2n + 1 = 10 k  +3 = 5 * 2k +3$ remainder is 3 which is a not a valid candidate

$n = 5k+ 2$ gives  $2n + 1 = 10 k  +5 = 5 * (2k +1)$ remainder is 0 which is a valid candidate

 $n = 5k+ 3 gives  $2n + 1 = 10 k  +7 = 5 * (2k +1)+2 $ remainder is 2 which is a not a valid candidate

$n = 5k+ 4$ gives  $2n + 1 = 10 k  +9 = 5 * (2k +1)+ 4 $ remainder is 4 which is a valid candidate

So valid candidates for 2n+1  being perfect square is 5k, 5k+2, 5k+ 4

and we need to check for these one 3n+1 is valid

we can apply above procedure to see  the$ n= 5k$ 3n+1 leaves remainder 1 which is valid and for $n=5k+2$ remainder is 2 and for $n=5k+4$ remainder is 3 both of which are invalid

so n is multiple of 5.

as n is divisible by 5 and 8 so by 40

2025/006) Let n be the smallest positive integer such that n is divisible by 20, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$

 Because $n^2$ is a perfect cube so n is a perfect cube

Because $n^3$ is a perfect square so n is a perfect square

so n is a perfect $6^{th}$ power

as n is divisible by 20$2^2 * 5}

so it is of the form $2^x5^y$ where x and y are multiple of 6

so $n = 2^6 * 5^6 = 10^6= 1000000$

number of digits in $n$ = 7

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

 $a^{12} \equiv 1 \pmod {35}$