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 $\{(552,115), (115,552), (345,232), (232,325)\}$