LHS
some short and selected math problems of different levels in random order I try to keep the ans simple
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}
= 2 \ sin \theta \cos 2 \theta + \sin \theta which is RHS
\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