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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2026/049) Does there exist integer solutions to $x^2+y^2=2022$

Before we look for solutions let us quickly check based on modular arithmetic.

 Looking at mod 3 we have 2022 is 2 mod 4. 

Solution may exist.

Let us check based on mod 3.

We have 

$2022 \equiv 0 \pmod 3$

But $2022 \equiv 6 \pmod 9$

So 2022 is not sum of 2 perfects squares

Let us prove the basis of the same

x is of the form $3a$ or $3a+1 or $3a+2$

If $x \equiv 0 \pmod 3$ then $x^2 \equiv 0 \pmod 3$

If $x \equiv 1 \pmod 3$ then $x^2 \equiv 1 \pmod 3$

If $x \equiv 2 \pmod 3$ then $x^2 \equiv 1 \pmod 3$

Similarly for y

Now looking at above we have $x^2+y^2 \equiv 0 \pmod 3 $ iff $x \equiv 0 \pmod 3$ and  $y \equiv 0 \pmod 3$.

So x = 3a and y = 3b for some a and b

or $x^2+y^2 = 9(a^2+b^2)$ or divisible by 9

But 2022 is not divisible by 9

Hence no solution

  

2026/048) For $1\le n \le2016$, how many integers n satisfying the condition: the remainder divided by 20 is smaller than the one divided by 16

Because LCM of 20 and 16 is 80. we need to look upto 80 an then the pattern repeats

Now remainder shall keep increasing till we get next multiple . then the remainder resets

Because we need to compare the remainder the number is of the form 20a + b and 16c + d ( 0 <=b <=19) , (o <=d <=16)

The remainder divided by 20 is lower than remainder divided by 16 if multiple of 16 is lower than multiple of 20. that is $16c < 20a < n < 16(c+1) $, That is the number lies between multiple of 20 and multiple of 16. The numbers that fit the condition are from 20a to 16(c+1)

Let us write multiples of 16 : $16,32,48,64,80$

multiple of 20 : -$20,40,60,80$

So the ranges are $0, 16(16*1),20(20*1),32(16*2),40(20*2),48(16*3),60(20*3),64(16*4),80$

the ranges where the remainder divided by 20 is lower in the rage $[20*1,16*2), [20*2,16 * 3]$ and $[20*3,16 * 4)$ in each range lower number is inclusive and higher is not giving $16*2 - 20 *1 + 16*3 - 20*2 + 16*4 - 20 *3$ or $24$ numbers.

In each lock of 80 numbers there are 24 numbers

upto 2000 there are 25 blocks so 25 * 24 600 numbers

From 2001 to 16 there are 16 number having same remainder as divided by 16 as divided 20

So no extra

So Ans 600