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.