Python Program for Consecutive Prime Sum (TCS Codevita) | PrepInsta

July 2, 2020

Python Program for consecutive prime sum

Consecutive Prime Sum Problem

Consecutive prime sum is one of the most popular challenging questions which was asked in TCS CodeVita Season 9 sample questions. This problems check your logical thinking ability. The difficulty level of this problem is between low-medium, regarding TCS CodeVita Season 9, other sample questions

All TCS CodeVita Questions with Answers

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

S.no Input Output Comment
1 20 2 (Below 20 there are two such members; 5 and 17) 5=2+3 17=2+3+65+7
2 15 1

Python Code

num = int(input())
arr = []
sum = 0
count = 0
if num > 1:
    for i in range(2, num + 2):
        for j in range(2, i):
            if i % j == 0:
                break
        else:
            arr.append(i)
def is_prime(sum):
    for i in range(2, (sum // 2) +2):
        if sum % i == 0:
            return False
        else:
            return True
for i in range(0, len(arr)):
    sum = sum + arr[i]
    if sum <= num:
        if is_prime(sum):
            count = count + 1

print(count)
Output
20
2

Consecutive Prime Sum Problem in Other Coding Languages

C

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C

C++

To find the solution of Consecutive Prime sum problem in C++ Programming language click on the button below:

C++

Java

To find the solution of Consecutive Prime sum problem in Java Programming language click on the button below:

Java

26 comments on “Python Program for Consecutive Prime Sum (TCS Codevita) | PrepInsta”


  • Meenal

    def check_prime(num):
    arr=[]
    count=0
    for i in (num):
    c=0
    for j in range(1,i+1):
    if(i%j ==0):
    c=c+1
    if(c==2):
    arr.append(i)
    s=arr[0]
    for j in range(1,len(arr)):
    s=s+arr[j]
    if(s in arr):
    count=count+1
    return (count)

    s=int(input())
    l=[]
    for i in range(1,s+1):
    l.append(i)
    p=check_prime(l)
    print(p)


  • Karthick

    end_num=int(input(“Enter the end number : “))
    prime=[]
    fin=[]
    for j in range(3,end_num+1):
    count=0
    for i in range(1,j):
    if(j%i==0):
    count+=1
    if(count==1):
    prime.append(j)
    summ=2
    while(summ<=prime[-1]):
    if(summ in prime):
    fin.append(summ)
    summ+=prime.pop(0)
    print(len(fin))


  • Vishal

    def something(arr):
    lastarr = []
    sum = arr[0]
    for i in range(1,len(arr)):
    sum = sum+arr[i]
    for j in range(1,len(arr)):
    if sum == arr[j]:
    lastarr.append(sum)
    return len(lastarr)

    num = int(input())
    arr = []
    sum = 0
    count = 0

    for i in range(2, num):
    for j in range(2, i):
    if i % j == 0:
    break
    else:
    arr.append(i)
    print(arr)
    ans=something(arr)
    print(ans)


  • Arjun

    def is_prime(n):
    c=0
    for i in range(1,n+1):
    if n%i==0:
    c+=1
    if c==2:
    return True
    else:
    return False
    m= int(input(‘enter limit’))
    input_list=[i for i in range(m+1)]
    li=list(filter(is_prime,input_list))
    s=5
    c=0
    for i in range(2,len(li)):
    if is_prime(s) and sm:
    break
    else:
    pass
    s=s+li[i]
    print(c)
    Hope it helps


  • Yash

    Yash Gurav Here,
    One of the best solution for the same problem is following :
    #Correct_Solution
    #list_manupulation
    #set_theory

    x=int(input())
    c=0
    b,d=[],[]
    e={2}
    for j in range(2,x+1):
    for i in range(2,j):
    if j%i==0:
    break
    else:
    b.append(j) #creaton of b
    for i in range(0,len(b)):
    c=c+b[i]
    d.append(c) #creation of d
    b_set=set(b)
    d_set=set(d)
    y=b_set.intersection(d_set)
    q=set(y)-set(e)
    print(len(q))


    • Anjali

      Please help me to understand what is wrong with this program
      n=int(input())

      a=[]
      for j in range (n):
      for i in range (2,int(j/2)):
      if (n%i)==0:
      break
      else:
      a.append(i)
      f=[]

      for m in range (len(a)-1):
      s=0
      for n in range (m):
      s=s+a[n]
      if (a[m]==s):
      f.append(a[m])
      else:
      break

      print(f)


  • Pankaj Kumar

    num = int(input())
    list = []
    count=0

    for i in range(2,num+2):
    for j in range(2,i):
    if i%j == 0:
    break
    else:
    list.append(i)
    sum=list[0]
    for j in range(1,len(list)):
    sum=sum+list[j]
    if sum in list:
    count+=1

    print(count)


  • Slayer166

    ashish
    def hcf1(n):
    for i in reversed(range(n)):
    if(n%i==0):
    hcf=i
    break
    return hcf
    n=int(input())
    sum1=2
    count=0
    for i in range(3,n+1):
    if(hcf1(i)==1):
    sum1+=i
    if(hcf1(sum1)==1 and sum1<n+1):
    count+=1
    print(count)


  • Sukesh

    #This is the correct the program

    Input=int(input()); List=[] ; ans=2; count=0; a=0; b=0
    for i in range(2,Input):
    if i ==2 or i==3 or i==5 or i==7: List.append(i)
    else:
    if i%2==1:
    if i%3!=0 and i%5!=0 and i%7!=0: List.append(i)

    if Input>5:
    for i in range(1,len(List)):
    a=List[i]
    ans+=a
    if ans<=List[-1]:
    if ans in List: print(ans);count+=1
    print(count)


  • Sukesh

    Input=int(input())
    List=[] ; ans=0; count=0; a=0; b=0
    for i in range(2,Input):
    if i ==2 or i==3 or i==5 or i==7:List.append(i)
    else:
    if i%2==1:
    if i%3!=0:
    if i%5!=0:
    if i%7!=0:
    List.append(i)
    if Input>5:
    for i in range(0,len(List),2):
    a=List[i]; b=List[i+1]
    ans+=(a+b)
    if ans<=List[-1]:
    if ans in List:
    count+=1
    print(count)


  • Aryan

    N=int(input())
    def isprime(N):
    for i in range(2,N):
    if(N%i==0):
    return(False)
    return(True)

    count=0
    for i in range(3,N):
    sum1=0
    if(isprime(i)==True):
    for j in range(2,i):
    if(isprime(j)==True):
    sum1+=j
    if(sum1==i):
    count+=1

    if(sum1>i):
    break
    print(count)


  • Vascular

    here is the efficient code

    num = int(input())
    arr = []
    sum = 0
    count = 0
    if num > 1:
    for i in range(2, num + 2):
    for j in range(2, i):
    if i % j == 0:
    break
    else:
    arr.append(i)
    def is_prime(sum):
    prime = [True for i in range(n+1)]
    l=[]
    p = 2
    while (p * p <= n):
    if (prime[p] == True):
    for i in range(p * p, n+1, p):
    prime[i] = False
    p += 1
    if prime[p]:
    return True
    else:
    return False

    for i in range(2, (sum // 2) +2):
    if sum % i == 0:
    return False
    else:
    return True
    for i in range(0, len(arr)):
    sum = sum + arr[i]
    if sum <= num:
    if is_prime(sum):
    count = count + 1

    print(count)


  • Archibrata

    import sympy
    # if you haven’t then instal pip sympy
    N = int(input())
    list_of_prime_numbers_within_input_range = [2,]
    output_list_containing_prime_sums = []

    # generating prime number using lambda function
    for x in range(3,n):
    if not any(y for y in range(2,1+int(x**(1/2))) if not x%y):
    list_of_prime_numbers_within_input_range.append(x)

    # or you could simply use sympy.prime(N)… Just an one line code

    for i in range(1,len(list_of_prime_numbers_within_input_range)+1):
    summ = sum(l[:i+1])
    if sympy.isprime(summ)== True and summ<=N:
    output_list_containing_prime_sums.append(summ)

    print(len(output_list_containing_prime_sums))

1 2