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Code for write a python program to find the prime numbers between 1 to 1000
def is_prime(n):
if n <= 1:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return True
def find_primes(n):
primes = []
for i in range(2, n+1):
if is_prime(i):
primes.append(i)
return primes
# test the function
print(find_primes(1000))
Output
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]
Explanation
The above solution has two functions:
is_prime(n)
takes an integern
as an input and returnsTrue
if it’s prime andFalse
otherwise. It uses a simple algorithm to check if the number is prime by iterating from 2 to square root of the number + 1 and check if the number mod any of them is 0 or not.find_primes(n)
takes an integern
as an input and returns a list of all prime numbers between 2 andn
(inclusive). It does this by using a for loop that iterates from 2 to n, for each number it calls theis_prime
function and checks if it’s prime or not. If it’s prime it appends it to the list
You can also use list comprehension to get the same output in a single line.
primes = [i for i in range(2,1001) if all(i % j != 0 for j in range(2, int(i ** 0.5) + 1))]
Find the compound interest from the given Principal, No. of Years, and Rate of Interest
Conclusion
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