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

 

 

 

 

2026/054) What is the next number of the sequence 256, 1156, 4356..

The sequence is the squares of numbers which leaves a remainder 56 when divided by 100 that is

$n \equiv 56 \pmod {100}$ 

We can look at the numbers mod 100 and then find the numbers from table lookup. However this is not interesting

To keep the things simple let us find the numbers mod 10 and then make it to numbers mod 100.

Let us look at the number mod 10 and find the square mod 10 we get (1st is number mod 10, 2nd is number square mod 10 $(0,0),(1,1),(2,4),(3,9),(4,6),(5,5),(6,6),(7,9),(8, 4),(9,1)$

So then umber is 4 mod 10 or 6 mod 10

That is number is of the form 10n+ 4 or 10n+ 6

Now we need to find n in both the forms.  

We can take the x as 2 digit number . adding 100 to x shall not change the 2 digits mod 100

This is because

 $(100n + k)^2 = 10000n^2 + 200nk + k^2$

so  $(100n + k)^2 = k^2\pmod {100}$

Consider first form $x = 10n + 4$

Now $(10n+4)^2  = 100n^2 + 80n + 16$

so $80n + 16 = 56\pmod {100}$

or $80n = 40 \mod 100$

dividing by 20 on both sides

$2n = 1\pmod 5$

or $n = 3 \mod 5$ or $n \in \{3,8\}$ as we are looking for single digit number

So the number is $34 \pmod {100}$ or $84 \pmod {100}$

we can combine both to get $x \equiv 34\pmod {50}$  

Considerfnext form $x = 10n + 6$

Now $(10n+6)^2  = 100n^2 + 120n + 36$

As remainder is 56 

so $20n + 16 = 56\pmod {100}$

or $20n = 20 \mod 100$

dividing by 20 on moth sided

$n = 1\pmod 5$

or $n = 1 \mod 5$ or $n \in \{1,6\}$ as we are looking for single digit number

So the number is $16 \pmod {100}$ or $66 \pmod {100}$

So we have  $x \equiv 16 \pmod {50}$ or $x \equiv 34 \pmod {50}$

  

 

 

 

 

 

 

 

2026/053) If both x and y are integers, how many pairs of solutions are there of the equation $(x−8)(x−10)=2^y$

As RHS is power of 2 so both $x-8$ and $ x-10$ should be power of 2 or they should be -ve of power or 2 because there are two terms on the left hand side so product shall be positive.

As $(x-8) - (x-10) = 2$ so the 2 values should be 4 and 2 or -2 and 4( because 2 and 4 the power of 2 which differ by 2)

Naturally x-8 shall be lower value

So taking $x -8 = 4$  or $x = 12$ we get $2^y= 4* 2 = 8$ or $y = 3$ giving $((x,y) = (12,3)$

Taking $x-8 = -2$ we get $x = 6$ and $y = 3$ or $((x,y) = (6,3)$

 So we have 2 set of solutions $(x,y) \in \{(12,3),(6,3)\}$

2026/052 ) Let $a, b, c, d$ be four integers. Prove that $(b−a)(c−a)(d−a)(d−c)(d−b)(c−b) $ is divisible by 12.

If is is divisible by 12 then it is divisible by 3 and also by 4. this is so because 3 and 4 are co-primes

First Let us prove that is divisible by 3

Let us take $(b-a) , (c-a), (d-a)$ mod 3

As taking mod 3 there are 3 different remainders there are 2 possibilities

1. One of them is zero so it is divisible by 3
2. Non of them is zero so at least 2 remainder are same say $b-a$ and $c-a$. then the difference $(c-a) - (b-a) or (c-b) gives remainder 0 so divisible by 3

Having proved divisible by 3 Now let us prove that is is divisible by 4

Let  us take $(b-a) , (c-a), (d-a)$

If at least two of them are even then it is divisible by 4

        1. If one is even say $d-a$ then $b-a$ and $c-a$ are odd and $(c-a) - (b-a)$ that is c-a is even . As                there are 2 even numbers so product is divisible by 4

        2. If none is even the $b-a$ and$ c-a$ being odd $c-b$ is even and $b-a$ and $d-a $being odd $d-b$             is even . As $c-b$  and  $d-b$ are even so product is divisible by 4

In all cases it is divisible by 4

Hence it is divisible by 12

2026/051) Show that square of sum of 3 squares can be expressed as sum of 3 squares

 We need to show that $(a^2+b^2+ c^2)^2$ can be expressed as sum of 3 squares.

Let for simplicity 

$x= a^2+b^2\cdots(1)$ 

and 

$y=c^2\cdots(2)$

So we have

$(a^2+b^2 + c^2)^2 = (x+y)^2$ from (1) and (2)

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

$=  (a^2+b^2-c^2)^2 + 4(a^2+b^2)c^2$ putting back values

$=  (a^2+b^2-c^2)^2 + 4a^2c^+4b^2c^2$

$=  (a^2+b^2-c^2)^2 + (2ac)^2 + (2bc)^2$

Hence proved  

 

2026/050) Show that if a,b,c are sides of a right angled triangle and c is the hypotenuse then $a^n+b^n \lt c^n$ for $c \gt 2$

We shall prove it by Principle of  Mathematical Induction

The base step is n = 3 and n = 4

Because  it is right angled triangle and c is the hypotenuse we have

$a^2 + b^2 = c^2\cdots(1)$

Further as  

$a < c$

Multiplying both sides by  a^2 we have

$a^3 < ca^2\cdots(2)$

Similarly 

$b^3 < cb^2\cdots(3)$

Adding (2) and (3) we get

$a^3 + b^3 < ca^2 + cb^2$

or 

$a^3+b^3 < c(a^2+b^2)$

or $a^3 + b^3 <c^3$ using (2)

Additionally

Squaring (1) we get 

$a^4 + b^4 + 2ab = c^4$

And hence dropping 2ab(which is positive) from the left 

$a^4 + b^4 < c^4$    

So we have proved the base step

For induction step we shall not move from n to n+1 but ftom n to n+ 2

We have Let t be true upto k = n

from k = n- 1  (not k) we shall prove for n+ 1

we have

$a^{k-1} < c^{k-1}$

Multiplying by a^2 on both sides we get

$a^{k +1} < c^{k-1}a^2\cdots(4)$

Similarly

 $b^{k +1} < c^{k-1}b^2\cdots(5)$

from (4) and (5) we get 

 $a^{k +1}  + b^{k+1} < c^{k-1}a^2 + c^{k-1}b^2$

Or  

 $a^{k +1}  + b^{k+1} < c^{k-1}(a^2 + b^2)$

or   $a^{k +1}  + b^{k+1} < c^{k-1}c^2)$

or  $a^{k +1}  + b^{k+1} < c^{k+1}$

So the induction step is complete

We have proved for 3 and 4 ad from n-1 to n+ 1 and hence for all above 2

As we have proved base step and induction step proof is complete

 

 

 

 

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