2026/020) From the number $7^{2026}$ we delete its first digit, and then add the same digit to the remaining number. This process continues until the left number has ten digits. Show that the left number has two same digits.

We are deleting one digit and adding to the number. So value mod 9 shall not change . Given number mod 9 is not zero . So when we get a 10 digit number it shall not be divisible by 9. If all 10 digits are different then the number is divisible by 9. So all ten digits cannot be different. So upto 9 digits shall be different so at least one digit must repeat or two same digits will be there.

2026/019) Determine all positive integers that are equal to 300 times the sum of their digits.

 

The number must end with 2 zeroes . Excluding the 2 zeroes the sum of digits multiplied by 3 is the number. It is not one digit number as multiplied by 3 of the digit changed number

Maximum sum of digits of a 3 digits number is 27. Multiplied by 3 gives 81 which is a 2 digits number . So number of digits cannot be 3 or more. So it is a 2 digit number if it exists. 

Let it be $10a +b$. 

We have 

$10a +b =3(a+b)$ 

Or $7a =2b$ 

As a and b are single digit number so a is 2 and b is 7 and number is 27. We have removed 2 zeroes and so number is 2700 .

 

 

2026/018) Let $lcm(a,b)$ denote the least common multiple of a and b. Find the sum of all positive integers x such that $x \le 100$ and $lcm(16,x)=16x.$

 x must be co-prime to 16 . That means x is odd and every odd number satisfies the criteria . So result  is sum of all odd numbers less than 100. That is 2500

2026/017) Let m and n be positive integers such that 5 divides $2^n+3^m$. Prove that 5 divides $2^m+3^n$

 

Because 5 is a small number we can work as below each exponent

Working in mod 5 we have

$,2^1 = 2,2^2= 4, 2^3 = 8 = 3,2^4=1$

$3^1=3, 3^2 = 9 = 4 , 3^3 = 27 = 2,3^4=1$

so $2^4 + 3^2$ = 0 mod 5 also $3^4 +2^2 = 0$ mod 5

$2^1+ 3^1 = 0$ mod 5 ( n=m)

$2^3 + 3^3 = 0$ mod 5(n=-m)

We have checked for all combinations that it is true

2026/016) Find all solutions of the linear congruence $3x−7y \equiv 11 \pmod {13}$

We have as 13 is a prime number we can choose any value of a x and the choose y in terms of x

We have $7y \equiv  3x -11  \pmod  {13}$

Now 7 needs to be multiplied by its inverse to get coefficient of y as 1 so multiplying by 2 (inverse of 7) we get

$y \equiv  6x - 22 \pmod  {13}$

Or   $y \equiv 6x-9  \pmod {13} $

For x = 0 to 12 mod 13 we get corresponding  value of y 

2026/015) Show that difference between squares of 2 conscutive triangular numbers is a perfect cube

We have $T_n = frac{n(n+1)}{2}$ 

To avoid fraction we have $2T_n = n(n+1)$

Now we have 

$ (2T_{n+1})^2 -   (2T_{n})^2 = ((n+1)(n+2))^2 -((n+1)(n+2))^2 $

Or $4(T_{n+1}^2-T_n^2) = (n+1)^2((n+2)^2 - n^2) = (n+1)^2 * 4 (n+1))= 4(n+1)^3$

Or   $T_{n+1}^2 -   T_{n}^2 = (n+1)^3$

 

2026/014) Find all integer N such that $\lfloor \sqrt{n} \rfloor$ divides n

 Any number from $k^2$ to $k(k +2)$ shall have square root between $k$ and $k+1$ as $k(k+2) +1$ shall have square root $k + 1$

So the floor function of any number $k^2$ to $k(k+2)$ shall be $k$. For $k$ to divide the number with in the range it has to be of the form $k^2$ or$ k(k+1)$ or $k(k+2)$

 So for any $k $ $k^2 , k(k+1),k(k+2)$ satisfy the criteria