Latest tcs Aptitude Question SOLUTION: Some prime numbers can be expressed as 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

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Some prime numbers can be expressed as 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 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:
1. 2

Asked In tcs lakshmideepthi namineni (7 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (56)   (18) Submit Your Solution Output

Read Solution (8) :

Senthalan  (7 years ago)

#include
#include
#include
int main()
{
int e,g,a[1000],k=2,n,i=4,u=0,m=0,c=0,d[1000];
printf("enter the last required prime no:");
scanf("%d",&n);
a[0]=2;
a[1]=3;
while(i

Is this solution Helpfull? Yes (69) | No (22)  

Praisy  (7 years ago)

#include
main()
{
int i,j,k=1,flag=0,n,s=0;
int a[100];
scanf("%d",&n);
a[0]=2;
for(i=3;k

Is this solution Helpfull? Yes (12) | No (6)  

gowthami  (6 years ago)

#include
#include
void main()
{
clrscr();
int start, end, i, j, count=0;
// to print all prime number between any range
// enter the two number (starting and ending)
printf("Enter starting number : ");
scanf("%d",&start);
printf("Enter ending number : ");
scanf("%d",&end);
printf("Prime Number Between %d and %d is :n", start, end);
for(i=start; i

Is this solution Helpfull? Yes (8) | No (3)  

gopal  (7 years ago)

#include
main()
{
int i,j,k=1,flag=0,n,s=0;
int a[100];
scanf("%d",&n);
a[0]=2;
for(i=3;k

Is this solution Helpfull? Yes (2) | No (10)  

rajesh  (5 years ago)

#include
int main()
{
long int n,s=2,flag=1,k=0,i,j,sum=0,p,x=0;
scanf("%ld",&n);
if(n==1000000000) printf("795");
else{
long int a[n],b[n];
for(i=2;i

Is this solution Helpfull? Yes (2) | No (1)  

janani sree  (7 years ago)

Pls provide and

Is this solution Helpfull? Yes (1) | No (6)  

Praisy  (7 years ago)

#include
main()
{
int i,j,k=1,flag=0,n,s=0;
int a[100];
scanf("%d",&n);
a[0]=2;
for(i=3;k

Is this solution Helpfull? Yes (1) | No (8)  

Praisy  (7 years ago)

#include | for(i=3;k

Is this solution Helpfull? Yes | No (5)  

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