2026/063) Factor $n^4+6n^3+11n^2+6n+1$

Let $f(n) =  n^4+6n^3+11n^2+6n+1$

This does not have a change of sign so there is no positive root

So it can have -ve root if it has real root it has to be -1

Checking using the rational root theorem $(f(-1) = 1 - 6 + 11 -6 +1 = 1$

So -1  is not a root

So it has to be product of 2 quadratic polynomials 

This is a quartic polynomal. And the coefficients are symmetric.

So if n is a root then $\frac{1}{n}$ is a root. So the coeffcient of$x^2$ and constant should same in both polynomials. However I shall continue as below

$f(n) =  n^4+6n^3+11n^2+6n+1$

$=  (n^4+1) +11n^2+6(n^3+n)$ reordering the terms

$= (n^2+1)^2 -2n^2 +11n^2+6n(n^2+1)$ putting $n^4+1$ in terms of $(n^2+1)$

$=(n^2+1)^2 + 6n(n^2+1) + 9n^2$

$= (n^2+1)^2+ 2(3n)(n^2+1) + (3n)^2$ geting expressing in $a^2+2ab+b^2$ form

$=(n^2+1+3n)^2$

$= (n^2+3n+1)^2$ putting in standard form  

 

2026/062) Define $f(n)=LCM(1,2,\cdots\,n)$. Determine the smallest positive integer a such that $f(a)=f(a+2)$

This shall require that $a+1$ and $a+2$ do not increase the LCM.

This shall require both $a+1$ and $a +2 $to be composite and neither is a perfect power of a prime. 

Let us explain it for $a+1$ and the same logic holds for $s+ 2$

If $a+1$ is prime then we have not encountered the same in any of the numbers and this shall contribute to LCM and it shall increase. LCM should be multiplied by $a+ 1$ 

If  $a+1 = b^k$ where b is a prime then it should be multiplied by $b$ as $b^(k-1)$ must have come previously but not $b^k$

if $a+1=b^kc^m$ that is product of powers of 2 primes then $b^k$ and $c^m$ has already come and hence it shall not contribute to  a higher LCM

Same for $a+2$     

 So we require smallest n such that n+1, n+ 2 are composite and producut of power of at least 2 primes 

Looking at $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$14 = 2 * 7 $so LCM does not increase

$15 = 3 * 5$ so LCM does not increase

For $(8,9)$ both power of primes ,

 For $(9,10)$, $9$ is power of prime

 So $a = 13$ 

Next $a = 19$ as  $20=2 ^2 * 5, 21 = 3 *7$ do not increase meets criteria  

  

2026/061) Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

Without loss of generality we can assume that nmbers are in increasing sequence 

We have 3 powers of 2 as 2 digit numbers . They are 16,32,64. 

If we start at 10 we get 4 numbers in geometric progression when ratio is 2. 

So we need some ratio less than 2 . Common ratio can be fraction as long as we do not get a fraction after multiplying by common ratio . 

Because we need 5 numbers we need to multiply 4 times . So we start with a 4th power of 2 that is 16 and common ratio $\frac{3}{2}$ 16 giving 16, 24,36,54,81 and middle term is 36.

Basically  5 numbers $a^4,a^3b,a^2b^2,ab^3,b^4$ where a is starting number and $\frac{b}{a}$ as common ratio. form  GP.

 

2026/060) Show that $f(n)=n^5+n^4+1$ is not prime for $n>1$

We shall prove a stronger result. we shall show that

$g(n) = n^{3k+2} + n^{3m+1} + 1\cdots(1)$ is not a prime for $n>1$ and $k+m > 0$

To prove it we shall snow that is is divisible by $n^2+n+1$

We have 

$n^2+n+1 = (n-\omega)(n-\omega^2)..\cdots$  where $\omega$ is cube root of 1

And $w^3 = 1\cdots(3)$

And $w^2+w+1=0\cdots(4)$ 

We shall snow that g(n) is divisible by  $ (n-\omega)$ and $ (n-\omega^2)$

 We get putting $\omega$ in (1) for n 

$g(\omega) =   \omega^{3k+2} + \omega^{3m+1} +1$

$= (\omega^3)^k \omega^2 +   (\omega^3)^m \omega + 1$

$=   \omega^2 +   \omega + 1 = 0$ using (3) and (4)

So $g(n)$ is divisible by  $(n-\omega)$

 Similarly $g(n)$ is divisible by $(n-\omega^2)$

So $g(n)$ is divisible by  $n^2+n+1$

Because  $k+m$ is greater than 0 so at least one of them is greater than zero

So $g(n) > n^2+n+1$ and as $n^2+n+1 > 0$ g(n) is product of 2 numbers neither is 1 

Hence g(n) is composite

 Putting k = m = 1 we get $f(n)$

So f(n) is not prime 

 Proved

 

2026/059) Show that the product of two positive integers of the form $a^2+ab+b^2$ has the same form.

Let two integers be $a^2+ab+b^2$ and $c^2+cd+d^2$

We have

$a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)\cdots(1)$ where $\omega$ is complex cube root of 1

And

$c^2+cd+b^2 = (c-d\omega)(c-d\omega^2)\cdots(2)$ 

 Multiplying (1) by (2) we get

 $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

Now as $\omega$ is cube root of 1 so we have

$\omega^2+\omega+1 = 0$

Or 

$\omega^2 = - (1+\omega)\cdot(3)$

Also

 $\omega = - (1+\omega^2)\cdot(4)$ 

Now get us calculate $(a-b\omega)(c-d\omega)$

$(a-b\omega)(c-d\omega) = ac -\omega(ad + bc) + bd\omega^2$

$= ac -\omega(ad+bc) -bd(1+\omega)$ from (3)

$= (ac-bd) -\omega(ad+bc+bd)\cdots(5)$

 

Now get us calculate $(a-b\omega^2)(c-d\omega^2)$

$(a-b\omega^2)(c-d\omega^2) = ac -\omega^2(ad + bc) + bd\omega^4$

$= ac -\omega^2(ad+bc) +bd(\omega)$ as $\omega^3=1$ 

$= (ac -\omega(ad+bc) +bd(1+\omega^2)$ from (4)

 $= (ac-bd) -\omega^2(ad+bc+bd)\cdots(6)$

Now   $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$

$=(a-b\omega)(c-d\omega)(a-b\omega^2)(c-d\omega^2)$

$=((ac-bd) -\omega(ad+bc+bd))((ac-bd) -\omega^2(ad+bc+bd))$

$=((ac-bd)^3 +(ac-bd)(ad+bc+db)+(ad+bc+db)^3$                               using $x^3+y^3=(x-\omega y)(x-\omega^2 y)$

= $m^2+mn + n^2$ where  $m= ac-bd$ and $n = ad+bc+bd$

Hence proved  

  

 

2026/058) Find the number of ordered triples of positive integers $(a,b,c)$ such that $6a+10b+15c=3000$

We are given

$6a+10b+15c=3000\cdots(1)$ 

We have rearranging the terms let us get a in terms of others 

$6a = 3000 - 10b-15c = 5(600-2b-2c)$

As the RHS is divisible by 5 and GCD(5,6) = 1 so a is divisible by 5 and so

$a=5x\cdots(2)$ for some x

Again to get b in terms of others

 $10b = 3000 - 6a-15c = 3(1000-2a-5c)$

As the RHS is divisible by 3 and GCD(3,10) = 1 so b is divisible by 3 and so

$b=3y\cdots(3)$ for some y

 Again to get c

 $15c = 3000 - 6a-10b = 2(1500-3a-5b)$

As the RHS is divisible by 2 and GCD(2,15) = 1 so c is divisible by 2 and so

$c=3z\cdots(4)$ for some z

Putting the values a,b,c from (2),(3),(4) respectively in (1)
 

$30x + 30y + 30z = 3000$

Or $x + y + z = 100$

We need to find x,y,z all 3 positive integers sum is 100

Let us assume that there are 100 stones in a line . There are 99 gaps. we can make them into 3 parts of each part containing above one by putting 2 sticks in 99 gaps. this can be done in $99 \choose 2$ ways

  

 

 

2026/057) Find the smallest positive integer n such that $3^5 | 10^n-1$

Before finding the n let us make a few obsevations

We have for any n $9 | 10^n-1$

That is   $3^2 | 10^n-1$

And as $ 3| 111$ or $3 | \frac{10^3-1}{9}$

Hence $3^3 | 10^3-1$

Now we need to move from $3^3$ to $3^5$

999 is divisible by 27

Appending another 999 to the left shall give 999999 and it is divisible by 27 appending 9 or 99 does not make if divisible by 27

But appending 999 to the left is adding $999 * 10^3$

As $243 = 27 *9$ we need to keep a-dding until the  

Let us keep appending say n terms

We get $999(1+ 10^3 + 10^6 + \cdots 10^{3n})$

For this to be divisible by 243 as 999 is divisible by 27  $1+ 10^3 + 10^6 + \cdots 10^{3n})$ must be divisible by 9.

Each term of   $1+ 10^3 + 10^6 + \cdots 10^{3n})$ divided by 9 leaves a remainder 1 so there should by 8 terms and so n= 24

Then we get the number = $999(1+10^3 + \cdots 10^24)$

$= (10^3-1)(1+10^3 + \cdots 10^24)= 10^27-1$

so n= 27