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Saturday, March 30, 2024

2024/021) Show that \sin 3\theta = \sin \theta + 2 \sin \theta \cos 2\theta and use it to prove \sin \frac{\pi}{9} + 2 \sin \frac{\pi}{9} \cos \frac{2\pi}{9} = \frac{\sqrt{3}}{2}

LHS

= \sin 3 \theta = \sin ( \theta + 2\theta) = \ sin \theta \cos 2 \theta + \cos \theta \sin 2\theta usng \sin(A+B) formula 

= \ sin \theta \cos 2 \theta + \cos \theta (2\sin \theta \cos \theta) using \sin 2\theta formula

= \ sin \theta (\cos 2 \theta + 2  \cos^2  \theta)

= \ sin \theta ( 2 \cos 2 \theta +  \cos 2  \theta + 1 ) using formula for \cos 2\theta

= \ sin \theta ( 2 \cos 2 \theta +   1 )

= 2 \ sin \theta  \cos 2 \theta +  \sin   \theta which is RHS

In the above putting \theta = \frac{\pi}{3}  we get

\sin \frac{\pi}{9} +  2 \sin \frac{\pi}{9} \cos \frac{2\pi}{9} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}

 

Sunday, March 24, 2024

2024/020) Show that there are infinitely many composite numbers of the form 10^n +3 (n = 1, 2, 3, ... ).

If n = 2 we have 10^2+3 = 103 this is divisible by 103 . 

Hence  10^2 + 3 = 0 \pmod {103}\cdots(1)

We know that 103 is a prime but we shall use this as a starting point to show that there are infinite numbers of the form 10^n+3 is divisible by 103.

As 103 is a prime so using Fermats little theorem

10^{102} = - 1 \pmod {103}

Squaring both sides we get

10^{204} =  1 \pmod {103}

So from (1) and above we have

10^{204n + 2} + 3=  0 \pmod {103}

except for n = 0 when the value is 103 all other values are above 103 and have a factor 103 so not a prime

So there are infinitely many composite numbers as they are divisible by 103. We have not taken all cases and have found infinitely many composite numbers. Hence proved

 


2024/019) Given a_n = 6^n + 8^n find the remainder when a_{83} is divided by 49

 We shall use Euler's theorem that is a generalization of Fermat's little theorem: For any n and any integer a coprime to n, one has 

a^{\phi(n)} \equiv -1 \pmod n

Where \phi(n) is number of numbers that is co-prime to n,

Let us compute \phi(49) 

If n = p_1^{k_1}p_2^{k_2}p_3^{k-3}  ... when each p_i is  prime then

 \phi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3})

As 49= 7^2 so \phi(49) = 49 * (1-\frac{1}{7}) = 49 * \frac{6}{7} = 42

Hence  

6^{\phi(49)} \equiv -1 \pmod {49}n

or 6^{(42)} \equiv -1 \pmod {49}

Squaring both sides

 6^{(84)} \equiv 1 \pmod {49}

Dividing by 6 we get

  6^{(83)} \equiv \frac{1}{6} \pmod {49}\cdots(1)

Where $\frac{1}{6}} is not fraction but inverse of 6 mod 49

Similarly

  8^{(83)} \equiv \frac{1}{8} \pmod {49}\cdots(2)

 Adding (1) and(2) we get 

  6^{(83)}+ 8^{(83)} \equiv \frac{1}{6}  + \frac{1}{8} \pmod {49}\cdots(2)

Now \frac{1}{6}  + \frac{1}{8} \pmod {49}

= \frac{8 + 6}{48} \pmod{49}

= \frac{14}{48} \pmod{49}

= \frac{14}{-1} \pmod{49} as 48 \equiv -1 \pmod {49}

= -14 \pmod{49}

=35 taking positive number 

Hence remainder = 35 

Sunday, March 17, 2024

2024/018) There are 1000 coins -- 999 are fair, and 1 has heads on both sides. You randomly choose a coin and flip it 10 times. Miraculously, all 10 flips turn up heads. What is the probability that you chose the unfair coin

Probability that you choose one unfair coin = \frac{1}{1000}

iI you choose one unfair coin the probability that all 10 heads come = 1

So   Probability that you choose one unfair coin and all heads come = \frac{1}{1000}

Probability that you choose one fair coin = \frac{999}{1000}

If you choose one fair  coin the probability that all 10 heads come = \frac{1}{2^{20}} = \frac{1}{1024}

So   Probability that you choose one fair coin and all heads come = \frac{999}{1024*1000}

 So probability that all heads come = \frac{999}{1024*1000} + \frac{1}{1000} = \frac{2023}{1024000}

 So probability that coin is unfiar = \frac{\frac{1}{1000}}{\frac{2023}{1024000}}=\frac{1024}{2023}

 

Saturday, March 2, 2024

2024/017) integrate sin ax cos bx

We have \sin \, ax + \cos \, bx = \frac{1}{2} (\sin (a+b) x + \sin (a-b) x

Knowing that 

\frac{d}{dx} \ cos \, nx = - n \sin \, nx

Hence \int  (\sin \, ax  \cos \, bx) dx  = \int (\frac{1}{2} (\sin (a+b) x + \sin (a-b) x)dx  

= \frac{\cos(a+b)x}{a+b} - \frac{\cos(a-b)x}{a-b} + C