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 

  

 

 

 

 

 

2026/056) Prove that 1280000401 is composite.

 We have 

$1280000401 = 1280000000 + 400 + 1= 20^7 + 20^2+ 1$

This is $x^7 + x^2 + 1$ where x = 20

All the terms have coefficient 1

There is an exponent 7 and and an exponent 2  one if of the form 3n + 2 and another of the form 3n+1

This is divisible by $x^2+x+1$ hence composite

Let us prove the same.

$x^7 + x^2 + 1$ 

$= x^7 -x + x^2 + x + 1$

$=x(x^6-1) + x^2 + x + 1$

$=x(x^3+1)(x^3-1) + (x^2 + x + 1)$

$=x (x^3+1)(x-1)(x^2+x+1)+ x^2 + x + 1$

$=(x^2+x+1) (x(x^3+1)(x-1) + 1$

We have factored the same and 

so the given  number is composite

 

 

 

 

2026/055) show that 1994 devides $10^{900} - 2^{1000}$

Now Let us factor 1994

$1994 = 2 * 997$

We need to show that it is divisible by 2 and 997 

RHS is even so is is divisible by 2

We need to show that the RHS is divisible by 997 

We have as 997 is close to 1000 so let is brrin power of 10 and 2 to be as close as 997

$10^3 \equiv 3 \pmod {997}\cdots(1)$

and  $2^{10} \equiv 27 \pmod {997}\cdots(2)$

so $10^{900} - 2^{1000}$

 $=({10^9})^{100} - (2^{10})^{100}$ )

This is divisible by $10^9 - 2^{10}$

Now $10^9 - 2^{10}\pmod {997}$

$= (10^3)^3 - 2^{10}\pmod {997}$

 $= 3^3 - 27 \pmod {997}$

$= 27-27 \pmod {997}$

= 0

So the number is divisible by 997

As it is divisible by 997 and 2 so it is divisible by 1994