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 

 

 

 

2026/047) Factorize: $2x^2−y^2−3z^2−xy+4yz+5zx$

To factor this we shall take it as this as a polynomial of  Let us write in decreasing power of x

$2x^2−y^2−3z^2−xy+4yz+5zx$

$=2x^2+x(-y+5z) -(y^2–4yz+3z^2)$

 $= 2x^2+x(-y+5z) -(y-z)(y-3z)$ by factoring the the term independent of z.

We let $a = y-z$ and $b= y-3z$ which are factors of the independent term and they are co prime to each other

 Now a and b are co-primes so 2 the coefficient of $x^2$ $must go with one of them and we have to chose the same so that the sum (or difference) is co-efficient of x.

We have $a - 2b = -y + 5z$

So we get 

$2x^2 +(a-2b) x - ab$

$= 2x^2 + ax - 2bx -ab$+

$= x(2x+a)-b(2x+ a)$

$= (x-b)(2x+a) = (x-y+3z)(2x+y-z)$

2026/046) Find the sum of the real roots of the polynomial $\prod_{k=1}^{100} (x^2-11x +k)$

 It is product of 100 terms. 

From each term we shall have 2 solutions. 

Either both the solutions are real or complex

Let check which and how many terms shall have real roots

$x^2 - 11x +k$ has real root if discriminant is positive or zero 

That is $11^2 - 4 * k \ge 0$ or $ k \le 30$

There are 30 terms for which it has got real roots and sum of real roots in each term is 30 giving sum of real roots 330  

2026/045) Prove that $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)=\tan 9^{\circ}$

We shall use formula for $\cos 3x$  

$\cos 3x = 4 \cos^3 x - 3 \cos x$

so we have $4 \cos ^2 x - 3 = \frac{\cos 3x}{ \cos x}$

Hence 

$4 \cos ^2 9^{\circ} - 3 = \frac{\cos 27^{\circ}}{\cos 9^{\circ}}$

And 

$4 \cos ^2 27^{\circ} - 3 = \frac{\cos 81^{\circ}}{\cos 27^{\circ}}$

so   $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)$ = $\frac{\cos 27^{\circ}}{\cos 9^{\circ}}$ *  $\frac{\cos 81^{\circ}}{\cos 27^{\circ}}$ 

= $\frac{\cos 81^{\circ}}{\cos 9^{\circ}}$

 = $\frac{\sin 9^{\circ}}{\cos 9^{\circ}}$ using $\cos \theta = \sin (90^{\circ}-\theta)$

 = $\tan 9^{\circ}$

Hence Proved  

 

2026/044) Each digit in the n-digit number N is 1. What is the smallest value of n for which N is divisible by 333,333?

We know 333333 = 3 * 111111

Now the number which is 111111 or $\frac{10^6-1}{9}$ 

Any number which is having  1's is $\frac{10^n-1}{9}$ 

 $\frac{10^6-1}{9}$ shall divide  $\frac{10^n-1}{9}$ only when n is multiple of 6

Or n = 6k for some integer k

Number is divisible by 111111 and 9 as GCD(111111, 9) = 333333 

This is so because 111111 is divisible by 3 and 9 = 3* 3

So 6k should be divisible by 9 and smallest k = 3 or 6k = 18. 

N = 6k = 18