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Java Loops | HackerRank solutions -codewithyasar

  Objective In this challenge, we're going to use loops to help us do some simple math. Task Given an integer,  , print its first   multiples. Each multiple   (where  ) should be printed on a new line in the form:  N x i = result . Input Format A single integer,  . Constraints Output Format Print   lines of output; each line   (where  ) contains the   of   in the form: N x i = result . Sample Input 2 Sample Output 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20 Explanation:           Here, we just need to use for loops to achieve the result Solution : import java.io.* ; import java.math.* ; import java.security.* ; import java.text.* ; import java.util.* ; import java.util.concurrent.* ; import java.util.regex.* ; public class Solution { public static void main ( String [] args ) t...
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Consecutive Prime Sum | Codevita |Python

Problem Description

Question – : Some prime numbers can be expressed as a sum of other consecutive prime numbers.
For example

5 = 2 + 3,
17 = 2 + 3 + 5 + 7,
41 = 2 + 3 + 5 + 7 + 11 + 13.
Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

Write code to find out the number of prime numbers that satisfy the above-mentioned property in a given range.

Input Format: First line contains a number N

Output Format: Print the total number of all such prime numbers which are less than or equal to N.

Constraints:
2<N<=12,000,000,000

Input
20 
Output:
2 (Below 20 there are two such members; 5 and 17) 5=2+3 17=2+3+65+7
Logic:
  •     Input the number
  •     Find the prime numbers upto range
  •     Find the sum of prime numbers consecutively 
  •     And also check If the sum is below the given number
  •     And also find count
  •     If the sum is above the given number Break
  •     print count
Program:
def isPrime(num):
if num < 2:
return False
for i in range (2, num):
if num % i == 0:
return False
return True
n = int(input())
primos = [i for i in range(1, n) if isPrime(i)]
count = 0
for i in range(len(primos)):
sum_ = sum(primos[:i+2])
if sum_<=n and isPrime(sum_):
count+= 1
if sum_>n:
break
print(count)
view raw ConsecutivePrimeSum.py hosted with ❤ by GitHub
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