The solution to Instagram Challenge: Hidden Logic Number Series!

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Sunday (18/08/2024), we posted a sequence challenge on our Instagram page! If you didn’t check the post kindly check and follow us on Instagram as well.

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The challenge was to decode the pattern in the following sequence: Given List: [4, 10, 27, 52, 125]

Today, we’re revealing the logic behind these numbers and giving you a complete solution. Let’s dive in!

The Hidden Pattern:

The sequence follows a unique pattern where each number is generated using the formula:

Formula: Prime Square + Index Number

Here’s how it works:

Prime Numbers: These are numbers greater than 1 that can only be divided by 1 and themselves. The first few prime numbers are 2, 3, 5, 7, 11, etc.
Square of a Prime Number: Multiply the prime number by itself. For example, the square of 2 is 4, and the square of 3 is 9.
Index Number: This is the position of the number in the sequence, starting from 0.

Breaking Down the Sequence:

4 = 2² + 0 (2 is the 1st prime, 2² = 4; Index = 0; 4 + 0 = 4)
10 = 3² + 1 (3 is the 2nd prime, 3² = 9; Index = 1; 9 + 1 = 10)
27 = 5² + 2 (5 is the 3rd prime, 5² = 25; Index = 2; 25 + 2 = 27)
52 = 7² + 3 (7 is the 4th prime, 7² = 49; Index = 3; 49 + 3 = 52)
125 = 11² + 4 (11 is the 5th prime, 11² = 121; Index = 4; 121 + 4 = 125)

Your Challenge: Find the Next 5 Numbers:

Now that you know the pattern, let’s extend the sequence by calculating the next 5 numbers using the same logic.

Solution in Different Programming Languages,

c programming Code:

#include <stdio.h>
#include <math.h>

// Optimized function to check if a number is prime
int is_prime(int n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  }
  return 1;
}

// Optimized function to find the next prime number
int next_prime(int n) {
  while (1) {
    n++;
    if (is_prime(n)) return n;
  }
}

int main() {
  int prime = 11;
  for (int i = 5; i < 10; i++) {
    prime = next_prime(prime);
    int next_number = prime * prime + i;
    printf("%d ", next_number);
  }
  return 0;
}

// Time Complexity: O(N * sqrt(N))
// Space Complexity: O(1)

Python Code:

import math

import math

# Optimized function to check if a number is prime
def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

# Optimized function to find the next prime number
def next_prime(n):
    while True:
        n += 1
        if is_prime(n):
            return n

prime = 11
for i in range(5, 10):
    prime = next_prime(prime)
    next_number = prime**2 + i
    print(next_number, end=" ")

# Time Complexity: O(N * sqrt(N))
# Space Complexity: O(1)

Java Code:

public class PrimeSequence {

    // Optimized function to check if a number is prime
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i * i <= n; i += 6) {
            if (n % i == 0 || n % (i + 2) == 0) return false;
        }
        return true;
    }

    // Optimized function to find the next prime number
    public static int nextPrime(int n) {
        while (true) {
            n++;
            if (isPrime(n)) return n;
        }
    }

    public static void main(String[] args) {
        int prime = 11;
        for (int i = 5; i < 10; i++) {
            prime = nextPrime(prime);
            int nextNumber = (prime * prime) + i;
            System.out.print(nextNumber + " ");
        }
    }
}

// Time Complexity: O(N * sqrt(N))
// Space Complexity: O(1)

Conclusion:

The challenge of decoding this sequence allows us to explore fundamental concepts in programming such as prime numbers, loops, and conditionals. By applying these concepts, we’ve unlocked the pattern and created a program that continues the sequence. Whether you’re coding in C, Python, or Java, understanding the underlying logic is key to solving problems like these.

Happy coding! 🚀

Next Steps

Try extending the sequence even further or modifying the pattern to create your own number sequences. The possibilities are endless!