Sum = 1 + 12 + 123 + 1234 + 12345 + … + 1234567891011…n how to sum%10007 mod answer

Question

test
input 2
output 13

Answer 1

thanks

Answer 2

You are welcome!

Answer 3

no input 3 output sum = 1+12+123=136%10007

Answer 4

import java.util.Scanner;
public class Series 
{
 public static void main(String args[]) 
 {
  Scanner sc = new Scanner(System.in);
  System.out.print("Enter the number of terms: ");
  int n = sc.nextInt();
  int s = 0, c;                       // s for terms of series, c for counter to generate n terms
  for (c = 1; c <= n; c++) {
   s = s * 10 + c;
   System.out.print(s + " ");
  }
 }
}

Output:
Enter the number of terms: 6
1,  12,  123,  1234,  12345,  123456.......

You just need to add up!😸

Answer 5

Here you go.
Sorry for the functions

https://code.sololearn.com/cK6EYzD4k9ub/?ref=app

Answer 6

I'm confused, please elaborate

Answer 7

Here is JavaScript:

https://code.sololearn.com/Ws9P3r5Q2MvK/?ref=app

Answer 8

Just add %10007

Easy!

I'll update my code right away.

Answer 9

public class series
{
public static void main(String args[])
{
int i,s=0;
for(i=1;i<=9;i++)
{
s+=(s*10+1);
}
System.out.println("The sum of the series:"+s);
}
}

Answer 10

Can you say it in java that works on blue j

Answer 11

The general term satisfies the recurrence relation a(k) = 10a(k-1) + k with a(1) = 1. You can solve for it by back-substituting in the previous terms and you should get a(k) = (10^(k+1) - 9k - 10)/81. Summing from k = 1 to k = n gives you the result you need, s(n) = (100(100^n - 1))/729 - (n(n+1))/18 - (10n)/81.

This is faster than calculating through a loop. Just ask for the input of n, plug it into the formula, and print the output of s(n).