2024/043) Given $x^2 = 2^ y + 2023$ find x+ y where x and y are natual numbers(IIT JEE 2024)

We are given

  $x^2 = 2^ y + 2023$

Now working mod 4 we have 

$2^ y + 2023 \equiv 2^y + 3 \pmod 4$

y cannot be greater than 1 as $2^y + 3 \equiv 3 \pmod 4$

As a square cannot be $3 \pmod 4$ so only possible value is y = 1 giving x= 45 and $x+y=46$

2024/042) For each positive integer n we consider the sequence of 2004 integers $\lfloor n+\sqrt{n}\rfloor ,\lfloor n+1+\sqrt{n+1}\rfloor ,\lfloor n+2+\sqrt{n+2}\rfloor,$ $\cdots,\lfloor n+2003+\sqrt{n+2003}\rfloor$. How do I find the smallest integer n, such that the 2004 numbers in the sequence are 2004 consecutive integers?

We know for any integer x

$\lfloor x+y \rfloor = x + \lfloor y \rfloor$

So we have

$\lfloor  n + k + \sqrt{n+k}  \rfloor = n + k  + \lfloor \sqrt{n+k} \rfloor$

From the given 2024 numbers by putting k = 0 t0 2023 we get

 $\lfloor  n +  \sqrt{n}  \rfloor = n +  \lfloor \sqrt{n} \rfloor$

 $\lfloor  n + 1 + \sqrt{n+1}  \rfloor = n + 1  + \lfloor \sqrt{n+1} \rfloor$

$\lfloor  n + 2 + \sqrt{n+2}  \rfloor = n + 2  + \lfloor \sqrt{n+2} \rfloor$

$\lfloor  n + 2023 + \sqrt{n+2023}  \rfloor = n + 2023  + \lfloor \sqrt{n+2023} \rfloor$

They are consecutive if we have

 $\lfloor \sqrt{n} \rfloor = \lfloor sqrt{n+1} \rfloor \cdots \lfloor \sqrt{n+ 2023} \rfloor$

as these are in increasing order we have

  $\lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+ 2023} \rfloor$

The 1st term should be as low as possible so n is a perfect square $x^2$

So we have $x = \lfloor \sqrt{x^2+ 2023} \rfloor$

 As we have the smallest $k=(x+1)^2$ such that $x <  \sqrt{k}$

so we have $x^2 + 2023 \lt x^2 + 2x + 1$ or $2024 \lt 2x$ so x = 1013 and

smallest $n = 1013^2= = 1026169$

 

2024/041) The GCD of 3 numbers is 30 and their LCM is 900. Two of the numbers are 60 and 150. What is the possible third number?

We have factoring all 4

$60 =  2^2 * 3 *5$

$150 = 2 * 3 * 5^2$

$30 = 2 * 3 * 5$

$900 = 2^2 * 3^2 * 5^2$

Because the  GCD is 30 so the number has to be multiple of 30.

Now let us consider highest power of 2 coming from the numbers 900 has 2 and 60 has so 3rd number need not have (might have but not necessary)

Consider highest power of 3 coming from the numbers 900 has 3 and 60 and 150 have 2 so 3rd number must have so it has to be multiple of $2 * 3^2 * 5$ that is 90 

Consider highest power of 5 coming from the numbers 900 has 2  and 150  has so 3rd number need not have (might have but not necessary)

So 3rd number has to be multiple of 90 and it has to de divisor of 900 say 90m

90m is a factor of 900 or  is a factor of 10 that is 1 or 2 or 5 or 10

giving 3rd number  one of 90,180,450,900

2024/040) The product of roots of equation $9x^2 - 18 |x| + 5 = 0$ is

 Above are roots of equation  $9x^2 - 18x + 5 = 0$ and  $9x^2 - 18x + 5 = 0$

the discriminant of of the equations are $18^2 - 4 * 9 * 5 = 324-180 = 144$

so roots are real and product of roots of  $9x^2 - 18x + 5 = 0$ is $\frac{5}{9}$ and same is the product of other equation and so we get overall product  $\frac{25}{81}$

2024/039) The sides of a right angled triangle are in AP and area of 24. What are the length of sides

 Let the sides be a, a-d and a + d

we have $(a-d)^2 + a^2 = (a-d)^2$

or $a^2-2ad + d^2 + a^2 = a^2 + 2ad + d^2$

or $a^2 -4ad=0$

 or $a(a-4d)=0$

so $a=4d$

area of the $\triangle$  is $\frac{a(a-d)}{2} = 24$

or $\frac{4d * 3d}{2} = 24$

or d= 2 and hence sides are 6,8, 12

2024/038) Solve for x $81^{\sin ^2x} + 81^{\cos^2x} = 30$

 we have $81^{\sin ^2x} + 81^{\cos^2x} = 30$

Or $81^{\sin ^2x} + 81^{1-\sin^2x} = 30$

or  $81^{\sin ^2x} + \frac{81}{81^{\sin^2x}} = 30$

Let  $81^{\sin ^2x}=y$

So we get $y + \frac{81}{y}=30$

or $y^2-30y + 81=0$

or $(y-3)(y-27)=0$

Or $y=3$ or $y=27$

$y=3$ mean  $81^{\sin ^2x}=3$ or $sin^2x = \frac{1}{4}$ or $sin^2x = \pm \frac{1}{2}$ or $x= \pi \pm \frac{n\pi}{6}$

$y=27$ mean  $81^{\sin ^2x}=27$ or $sin^2x = \frac{3}{4}$ or $sin^2x = \pm \frac{\sqrt{3}}{2}$ or $x=  \pi \pm \frac{n\pi}{3}$

hence combining both we get $x= \frac{n\pi}{2}\pm \frac{\pi}{6}$

2024/037) How do I find some natural number a, such that 2a is a perfect square, 3a is a perfect cube, 5a is the fifth power of some natural number?

 The number is of the form $2^x3^y5^zm^{30}$ because the it has 2 ,3 and 5 as factors and  $m^{30}$ is a $30^{th}$ power because it is a square, cibe and fith power so multiplying by this shall satisfy the criteria.

$2^{x+1}$ has to be perfect square and $2^{x}$ is a  perfect cube and a  fifth power.

so $x+1 \equiv 0 \pmod 2\cdots(1)$

$x \equiv 0 \pmod 3\cdots(2)$

$x \equiv 0 \pmod 5\cdots(3)$

we can solve the same by chinese remainder theorem but   we use a short method

From (2) and (3)

$x \equiv 0 \pmod {15} \cdots(4)$

from (1) and (4) we have checking multiple of 15 that is 0 and 15

$x \equiv 15 \pmod {30} \cdots(5)$

similarly we evaluate y

 $3^{y+1}$ has to be perfect cube and $3^{x}$ is a  perfect square and a  fifth power.

so $y+1 \equiv 0 \pmod 3\cdots(6)$

$y \equiv 0 \pmod 2\cdots(7)$

$y \equiv 0 \pmod 5\cdots(8)$

From (7) and (8)

$y \equiv 0 \pmod {10} \cdots(9)$

from (6) and (9) we have checking mutiple of 10 that is 0,10,20

$y \equiv 20 \pmod {30} \cdots(10)$

 similarly we evaluate z

 $5^{z+1}$ has to be perfect fifth cube and $5^{x}$ is a  perfect square and a  perfect cube.

so $z+1 \equiv 0 \pmod 5\cdots(11)$

$z \equiv 0 \pmod 2\cdots(12)$

$z \equiv 0 \pmod 3\cdots(13)$

From (12) and (13)

$z \equiv 0 \pmod {6} \cdots(14)$

from (11) and (14) we have checking mutiple of 6 that is 0,6,12,18,24

$z \equiv 24 \pmod {30} \cdots(10)$

we can take principal vaues and get

$a=2^{15}3^{20}5^{24}m^{30}$