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

2016/013) Show that the tens digit of $3^n$ is always even

We know working in modulo 20 as 3 is co-prime to 20 

$3^0  \equiv 1 \pmod {20}$

$3^1  \equiv 3 \pmod {20}$

$3^2  \equiv 9 \pmod {20}$

$3^3  \equiv 7 \pmod {20}$

$3^4  \equiv 1 \pmod {20}$

It repeats from this point,

As $3^n \pmod {20}$  is single digit so tens digit is even.

2026/012) The sum of two numbers is 667, the ratio of their LCM to GCF is 120. What are the two numbers?

 

Let $GCD(m,n) = k$ then we have

$m = ka$

$n = kb$

For some integers a and b where GCD(a,b) = 1

LCM = kab and GCD = k

So $ab = 120$

And $m +n = k(a+b)$

$m +n = 667 = 29 * 23$

So k can be $1,23,29, 667$

If $k =$1$  $a +b = 667$ and $ab = 120$ this is not possible

If $k = 667$ $a + b = 1$ so this is not possible

If $k = 23$ $a +b = 29, ab = 120$ so $a = 24,b =5$ or vice versa giving $(552,115)$  or $(115,552)$

if $k = 29$ $a +b = 23, ab = 120$ so $a = 15, b =8$ or vice versa giving$ (345,232)$ or $(232,345)$

So solution set is $\{(552,115), (115,552), (345,232), (232,325)\}$

2026/011) Solve in integer $x^2+4x +2 \equiv 0 \pmod 7$

 To complete square add 2 on both sides to get 

$x^2+4x +4 \equiv 2 \pmod 7$

or $(x+2)^2  \equiv 2 \pmod 7\cdots(1)$

now working mod 7

 $(0^2  \equiv 0 \pmod 7\cdots(2)$

 $(1^2  \equiv 1 \pmod 7\cdots(3)$

 $(2^2  \equiv 4 \pmod 7\cdots(4)$

 $(3^2  \equiv 2 \pmod 7\cdots(5)$ 

 from  (1) and (5) we have

 $(x+2)  \equiv 3 \pmod 7\cdots(6)$ or  $(x+2)  \equiv 3 \pmod 7\cdots(7)$ 

or $x\equiv 1 \pmod 7$ or    $x\equiv 2 \pmod 7$

 

 

 

2026/010) Is it true that $a^2+ab+b^2$ is divisible by 7 for infinitely many coprime pairs (a,b)

As a = 0 gives b= 0 so a = 7k and b= 7m satisfy the given equation but they ate not co-prime  

Because of symmetry let is assume $a=mb \pmod 7$ 

Putting in the given equation we get $m^2+m+1 \equiv 0\pmod 7$

As $m=1$  is not a solution to it multiplying by (m-1) on both sides   we get 

$m^3 \equiv 1 \pmod 7$ 

So $ m \equiv 1  \pmod 7$ or $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$ but as $ m \equiv 1  \pmod 7$ is not a root of original eqution so solution is $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$

So we can chose $(7p+m, 7q+2m)$ or $(7p+m, 7q+2m)$  p and q to be chosen such that the pairs form a co-prime

One set the infinite co-primes $(7p+1,7p+2)$ for any p 

 

 

 

 

2026/009) If $abc=1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ how can I show that a or b or c equals 1

We have $abc=1\cdots(1)$

 So $\frac{1}{a} = \frac{abc}{a} = bc$

Similarly 

$\frac{1}{b} = ca$

And 

$\frac{1}{c} =   ab$

Hence

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = bc+ca+ab\cdots(2)$

From above and

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

We have  

Hence $a+b+c = ab+bc+ca\cdots(3)$ 

a,b c are roots of equation

$P(x) = x^3-(a+b+c)x^2 +  (ab+bc+ca)x - abc=0\cdots(4)$

Let $a + b+c = m\cdots(5)$

So we get  from (4), (3), (5) 

$P(x) = x^3-mx^2+mx -1=0$ 

1 is a root of above equation as P(1) is zero

As a , b,c are roots one of $a,b,c$ is 1

Proved  

2025/008) Is it true that every odd number divides some number of the form $2^n−1$

 

Let the odd number be k. now let us take the number sequence $2^2–1, 2^3–1, 2^4–1$ so on k values that is upto $2^k-1$

Let us calculate all k remainder values when the above expressions are divided by k (as the remainder can be from zero to k-1) then we have on difference is zero and we are done.

If all k values are not different then we have say for 2 values of m and n the remainder must be same. without loss of generality let us assume $m > n$

So $(2^m-1) - (2^n-1) = 2^(m-n) -1 * 2^n$ is divisible by k. As k is odd $gcd(2^n,k) = 1$ so k must divide $2^(m-n) -1$

Note: The above is for the persons who are not familiar with number theory  or want to know from $1^{st}$ principle. 

n divides $2^{\phi(n)} -1$ as n is co-prime to 2 as per  Eulers theorem in number theory.

2025/007) Can two different primes $p\ne q$ satisfy $p^q−q^p=1$

Because difference is one one of the term is even and other is odd . So either p or q (being prime) is 2.

Let p = 2

so $2^q-q^2$, $q =3$ gives $-1$,$ q = 5$ gives $7$ and this keeps going increasing so no solution

Now let us check with $q = 2$ 

So $ p^2 -2^p$  $p =3$ satisfies$ p = 5$ gives $-7$ and larger p gives smaller value decreasing so no more solution

So only solution $p =3, q = 2$

2026/006) Prove that for real $x_1,x_2,x_3$ $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$

We have

 $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)$

$=(x_1^2- \frac{4}{3}x_1x_2) + ( x_3^2-  \frac{4}{3} x_2x_3)+x_2^2$  Combining $x_1x_2$ with $x_1^2$ term and   $x_3x_2$ with $x_2^2$ term 

$=(x_1^2- 2 *\frac{2}{3}x_1x_2 + (\frac{2}{3}x_2)^2) + ( x_3^2-  2* ( \frac{2}{3} x_2x_3+  (\frac{2}{3}x_2)^2)+(x_2^2-  (\frac{2}{3}x_2)^2) - (\frac{2}{3}x_2)^2)$  completing the square and subtracting the same 

$=(x_1 - \frac{2}{3}x_2)^2 + ( x_3- \frac{2}{3}x_2)^2+\frac{1}{9} x_2^2$

This is positive or zero

So 

  $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)>=  0$

Hence  $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$ 

2026/005) How do I the smallest four-digit integer N, such that $\sqrt{3\sqrt{N}}$ is an integer?

We have one 3 under the square root so we should have another 3 that should be part of $\sqrt{N}$ So it should be $N= (3x)^2$ and x has to be a square so smallest 4 digit number giving $N= 9x^4$ satisfying that it is smallest 4 digit number    when $x=4$ giving  $N= 9 * 256$ or 2304 

2026/004) If roots of the equation $a(b−c)x^2+b(c−a)x+c(a−b)=0$ are equal, then prove that 'a', 'b', 'c' are in harmonic proportion.

 Because roots are equal so the descrminamt is zero

or $b^2(c-a)^2-4ac(b-c)(a-b) = 0$

becuase we need to show that $a,b,c$ are in HP so $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP

now given relation is

$\frac{1}{q^2} (\frac{1}{r} - \frac{1}{p})^2- 4(\frac{1}{p})(\frac{1}{q})(\frac{1}{q}- \frac{1}{r}) (\frac{1}{p}- \frac{1}{q})=0 $ 

or $\frac{(p-r)^2}{p^2q^2r^2}$$ - 4(\frac{(r-q)(q-p)}{q^2r^2p^2})=0 $

or $(r-p)^2 -4 (r-q)(q-p) = 0$

Let $x= r-q$ and $y = q-p$

so we get $r - p = x + y$

or $(x+y)^2 - 4xy= 0$

or $(x-y)^2 = 0$

or $x = y$ or $q-p = r-q$ so   $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP and hence $a,b,c$ are in HP 

 

 

2026/003) Solve in positive integers $a^2=b^4 + 16b + 1$

 b has to even because if b is odd then we have RHS is of the form 4k +2 and it cannot be a perfects squares

now next square after $b^4$ is $(b^2+1)^2$ and the given expression can be a square(not necessary it will be) if greater than or equal to $(b^2+1)^2$

$b^4 + 16b + 1 >= (b^2+1)^2$

or $b^4 + 16b + 1 >= b^4 + 2b^2 +1 $

or $16b >= 2b^2$ or $ b <= 8$

So we need to try even values of b less or equal to 8 giving $b= 2 =>  a^2 = 49 =>a = 7$ 

$b =4 => a^2 = 256 + 64 +1 = 321$ no solution

$b = 6=> a^2 = 1395$ no solution

$b = 8 => a = b^2+1 = 65$

Solution $(7,2)$ and $(65,8)$

2026/002) Let a,b,c be integers. If $4a+5b−3c$ is divisible by $19$, prove that $ 6a−2b+5c$ is also divisible by $19$

 As 4 and 6 have LCM 12 so we should multiply by 3 and proceed

$4a + 5b -3c$ is divisible by 19 so multiply by 3 to get $12a + 15b -9c$ is divisible by 19. adding $19c-19b$ we get

$12a - 4b +10c$ is divisible by 19

or $2*(6a-2b + 5c)$ is divisible by 19 and as 2 is co-prime to 19 hence $6a-2b+5c$  is divisible by 19

2026/001) How do I find the smallest positive integer n, such that each digit of n is either 0 or 7, and n is divisible by 15?

It is divisible by 15. So it is divisible by both 3 and 5

Because it is  divisible by 5 so unit digit is 0 or 5, As we have to chose from 0 and 7 so it is zero. Now we cannot have a zero in other position as it will increase the value of the number as it is additional digit and does not serve any other purpose.Sso other digits shall be 7 and we require 3 7s to be divisible by 3 or the number as 7770( 15 * 518) .