Mathematics

Back to school level Maths :)

Maths Basics-I

- Prime Numbers
- Divisors
- Prime Factorisation
Prime Numbers
Divisors?
Prime Factorisation?
Maths Basics-II

- GCD
- Euclid's Algorithm
- LCM brute force
- LCM using GCD
- GCD of N Numbers
GCD Brute Force
Euclid's Algorithm
GCD's of N Numbers
LCM Brute Force
LCM using GCD
Mathematics
continued....

Maths Basics-III

- Fibonacci Series
- Arithmetic Progression
- Geometric Progression
Trailing Zeroes

- Find the number of trailing zeroes in the value of N! where N is input, without finding the factorial as N! can get big.
Fibonacci Number

Write a program to compute the nth Fibonacci Number.
Fibonacci Series:
0, 1, 1, 2, 3, 5, 8 , ........
Arithmetic Progression

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.
Arithmetic Progression

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.
For example, 1, 5, 9, 13, 17, 21, 25, 29, 33, ...
a = 1 (the first term)
d = 4 (the "common difference" between terms)
AP Nth Term

For example, 1, 5, 9, 13, 17, 21, 25, 29, 33, ...
a = 1 (the first term)
d = 4 (the "common difference" between terms)
an = a + (n-1)d
Nth term of an AP
Arithmetic Progression

For example, 1, 5, 9, 13, 17, 21, 25, 29, 33, ...
a = 1 (the first term)
d = 4 (the "common difference" between terms)
Sn = n(2a + (n-1)d)/2
Sum of N terms of an AP
AP Sum of N Terms

1, 5, 9, 13, 17, 21, 25, 29, 33, ...
a = 1 (the first term)
d = 4 (the "common difference" between terms)
Sn = (n/2)*(2a + (n-1)d)
Sum of N terms of an AP
Geometric Progression

A geometric progression is a progression where every term bears a constant ratio to its preceding term.

GP Nth Term

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Maths Basics-IV

- Binomial Coefficients
- Permutation
- Combination
Combinatorics

Combinatorics is all about number of ways of choosing some objects out of a collection and/or number of ways of their arrangement.
Real Example
For example suppose there are five members in a club, let's say there names are A, B, C, D, and E, and one of them is to be chosen as the coordinator.
Clearly any one out of them can be chosen so there are 5 ways. Now suppose two members are to be chosen for the position of coordinator and co-coordinator. What is the number of ways to choose the two?
Permutation
Let's generalize it. Permutations of choosing R distinct objects out of a collection of N objects can be calculated using the following formula:

Combination
Combinations of choosing R distinct objects out of a collection of N objects can be calculated using the following formula

Basic Combinatorics Rules:
Suppose there are two sets A and B with finite elements.
-
The Rule of Product
The product rule states that if there are X number of ways to choose one element from A and Y number of ways to choose one element from B, then there will be X×Y number of ways to choose two elements, one from A and one from B. -
The Rule of Sum
The sum rule states that if there are X number of ways to choose one element from A and Y number of ways to choose one element from B, then there will be X+Y number of ways to choose one element that can belong to either A or to B.
Binomial Coefficient 👨💻

Choosing 3 Cats out of 5 Cats?
Maths Basics-V

- Logarithm
- Wrapper Classes*
- Big Integer in Java
- Math Module

[Java 08] Maths
By Prateek Narang
[Java 08] Maths
- 8