2024/016) Prove that product of 4 consecutive positive integers cannot be a perfect cube.

Proof;

We  shall use the fact that if x and y are co-prime then xy is a perfect cube if both  x and y are perfect cubes.

Let the 1st number be x so the product is $x(x+1)(x+2)(x+3)$

Either x is odd or even

If x is odd then we have $GCD(x,x+2) = GCD(x+1, x+2) = GCD(x+2,x+3) = 1$

For x + 2 to be a perfect cube minimum x is 6 

So $GCD(x+2,x(+1)(x+2)) = 1$

So we need to prove that $x(x+1)(x+3)$ is not a perfect cube.

$x(x+1)(x+3) = x^3 + 4x^2 + 3$

We have $x(x+3) - (x-1)^2 = x^3+3x - x^2 -2x -1=  x-1 > 0$ as x is minimum 6 

So $x(x+1)(x+3) >= (x+1)^3$

  $x^3 + 4x^2 + 3$ is below  $(x+2)^3 = x^3 + 6x^2 + 12x + 8$

As it is between 2 cubes it cannot be a cube

Similarly we can prove taking $x +1$ when $x$ is even

 

2024/015) Find natual number n such that $2^n + n | 8^n + n $

 1,2,4,6 

We know that $ x+ y | x^3+y^3 $

Hence $2^n + n | (2^3)^n + n^3$

as $2^n + n | 8^n + n $

so $2^n + n | n^3-n  $

so we must have $ n^3-n =0 $ or $2^n + n <  n^3-n  $

$n^3-n= 0$ gives n = -1,0, 1 and out of theses only 1 is solution

we need to solve  $2^n + n <  n^3-n  $ or  $2^n  <  n^3-2n  $

let us find an upper bound for n putting a condition 

$2^n < n^3$ for  $n \lt 10$

putting n from 1 to 9 we see that $n \in \{1, 2,4,6\} $ satisfy the case and there is no other solution                                                                                                                                               

 

 

2024/014) Solve $x^2-y=111$ and $y^2-x=111$ for $ x\ne y$

 We are given 

 $x^2-y=111\cdots(1)$

 $y^2-x=111\cdots(2)$

From (1) and (2)

$x^2-y = y^2 -x$

Or $x^2-y^2 + x -y = 0$

Or $(x-y)(x+y) + (x-y) = 0$

Or  $(x-y)(x+y+1) = 0$

As  $ x\ne y$  dividing by x-y we get $x+y+1=0\cdots(3)$

Or $y = -(x+1)$

Putting the value in (1) we get $x^2 + (x+ 1) = 111$ or $x^2 + x - 110 = 0$

Or $(x-10)(x+11) = 0 $

Or $ x= 10$ or $x = -11$

Putting in (3) if $x= 10$ then $y = -11$

If $x= -11 $ then $y = 10$

So Solution set $x=10,y= -11$ or $x=-11,y=10$

2024/013) Prove that $123123 | 2^{60}-1$

To show that 123123 is a factor of $2^{60}-1$ we need to show that each prime factor of 123123 to the highest power is a factor

Now let us factor 123123

$123123= 123 * 1001 = 3 * 41 * 7 * 11 * 13$ 

As each prime occurs once we need to prove each of 3,7,11,13,41 divides  $2^{60}-1$

We know $x^y-1 | x^{my}-1$ for any m

We have $2^2-1 |  2^{60}-1$  and $3| 2^{2} -1$ using FLT(Fermat's Little Theorem)  so $3| 2{60}^-1$

$2^6-1 |  2^{60}-1$  and $7| 2^{6} -1$ using FLT(Fermat's Little Theorem)  so $7| 2{60}^-1$

$2^{10}-1 |  2^{60}-1$  and $11| 2^{10} -1$ using FLT(Fermat's Little Theorem)  so $11| 2{60}^-1$

 $2^{12}-1 |  2^{60}-1$  and $13| 2^{12} -1$ using FLT(Fermat's Little Theorem)  so $13| 2{60}^-1$

Now we need to prove for 41

using FLT we know  $41| 2^{40} -1$

So let us find $GCD(2^{(60}-1,2^{40}-1)$

$GCD(2^{(60}-1,2^{40}-1)= GCD((2^{(60}-1) - (2^{40-1},2^{40}-1)$

$=GCD((2^{60} - 2^{40}), 2^{40}-1)$

$=GCD((2^{40}( 2^{20}-1), 2^{40}-1)$

 $=GCD((2^{20}-1), 2^{40}-1)$ as second number is odd so removing even factors of 1st number

$= 2^{20}-1$ as this is a factor of  $ 2^{40}-1$

now $= 2^{20}-1)= (2^{10}-1)(2^{10}+1)$

$=  1023 * 1025$ as $41 | 1025$ so $41 | 2^{60}-1 $

as each of 3,7,11,13,41 divides  $2^{60}-1$ so 123123 is a factor

2024/012) Given $a+b+c$ is divisible by 6 prove that $a^3+b^3+c^3$ is divisible by 6

We have $a^3-a= a(a^2-1) = a(a-1)(a+1) = (a-1)a(a+1)$ 

As $(a-1)a(a+1)$ is product of 3 consecutive  if is s divisible by 6 say 6m for some m

So $a^3 = a + 6m\cdots(1)$

Similarly 

$b^3 = b + 6p\cdots(2)$

and

$c^3 = c + 6q\cdots(3)$

Adding (1),(2) and (3) we get $a^3+b^3+c^3 = (a+b+c) +6(m+p+q)$

So if   $(a+b+c)$ is divisible by 6 then $a^3+b^3+c^3$ is divisible by 6

2024/011) Find all values of n such that $6^n+1$ has all digits same.

We have $6^2=36$ that is it ends with 6.

So we have all the powers of 6 end with 6

So $6^n+1$ shall end with 7

So we need to find n such that all the digits of $6^n+1$ must have all digits 7 and let it be k 7's/

So $6^n+1 = \frac{7}{9} (10^k -1 )$

Or $9(6^n+1) = 7 (10^k -1 )$for

Or  $9 * 6^n + 16 = 7 * 10^k\cdots(1)$

We have $2 | 6$ so  $2^2 | 6^2 $ so if $n \ge  5$ then we have  

$9 * 6^n + 16 \equiv 16 \pmod {32} $ for $n \ge  5$

So   $9 * 6^n + 16 \equiv m  \pmod {32} $ where $m \le 16$ and not zero

For $k \ge 5$ $ 7 * 10^5 \equiv 0  \pmod {32} $

So $k \le 4$ 

So we need to check for n such that  $9 * 6^n + 16 \le = 70000$

Or $6^n \le  7776$

Or $6^n+1 \le 7777$

We calculate for n = 1 $6^1 + 1 = 7$ meets criteria

 n = 2 $6^2 + 1 = 37$ does not meet criteria

n = 3 $6^3 + 1 = 217$ does not meet criteria

n =4  $6^4 + 1 = 1297$ does not meet criteria

n = 5 $6^5 + 1 = 7777 $  meets criteria

Other value of n takes the value out of range

So $ n \in \{1,5\}$

 

 

 

 

 

 

2024/010) Prove that $18 ! \equiv -1 \pmod {437} $

 We first factorize 437

$437  = 19 * 23$

Now let compute mod relative to19 and 23

As 19 is prime number as as per Wilson'sTheorem we have

  $18 ! \equiv -1 \pmod {19}\cdots(1) $

As 23 is prime number as as per Wilson'sTheorem we have

  $22 ! \equiv -1 \pmod {23} $

Now  

$22  \equiv -1 \pmod {23}\cdots(2) $

$21  \equiv -2 \pmod {23} \cdots(3)$

$20  \equiv -3 \pmod {23}\cdots(4) $

$19  \equiv -4 \pmod {23}\cdots(5) $

As $22! = 22 * 21 * 20 * 19 * 18! $

So $22 ! \equiv -1 \pmod {23} $

$\implies  22 * 21 * 20 *  19 *18 ! \equiv -1 \pmod {23} $

 $\implies  (-1) *(-2) * (-3) * (-4) *18 ! \equiv -1 \pmod {23} $

 $\implies  24 *18 ! \equiv -1 \pmod {23} $

 $\implies  24 *18 ! \equiv -1 \pmod {23} $

$\implies  1 *18 ! \equiv -1 \pmod {23} $

$\implies  18 ! \equiv -1 \pmod {23}\cdots(6) $

Using (1) and (3) we get 

  $18 ! \equiv -1 \pmod {437} $

Proved

2024/009) Find $\sqrt{\sqrt{9} - \sqrt{8}}$

We have $8 = 2 * 4 = 2 * 2^2$

so    $\sqrt{8} = 2 \sqrt{2}$

now $\sqrt{9} = 3$

so   $\sqrt{\sqrt{9} - \sqrt{8}}$

=  $\sqrt{3 - 2\sqrt{2}}$

this is of the form  $\sqrt{n - 2\sqrt{n-1}}$ when n = 3

so  $\sqrt{3 - 2\sqrt{2}} = \sqrt{2 + 1 - 2 * \sqrt{2} * 1}$

$= \sqrt{(\sqrt{2})^2 + 1^2 - 2 * \sqrt{2} * 1} $

$= \sqrt{2} - 1$