To compute this sum, we can use a nifty algebraic approach:
Let S be the sum, so:
S = 3 + 6 + 9 + 12 + ... + 996 + 999
We can also write S as:
S = 999 + 996 + ... + 12 + 9 + 6 + 3
Adding these two equations together we get:
2S = 1002 + 1002 + ... + 1002 + 1002
Where there are 333 1002's in the above equation.
This leads us to:
2S = 1002 * 333
S = 501 * 333 = 166,833.
The numbers divisible by all five numbers must be multiples of the LCM (LCD) of the numbers, which is 22 x 3 x 5 = 60. There are 5 numbers: 240, 300, 360, 420, and 480. --- There is also a C program that can be used to find these: #include <stdio.h> void main() { for (int i=200;i<=500;++i) if (!((i%6)(i%4)(i%5)(i%3)(i%2))) printf("%d ",i); printf("\n"); }
#include<stdio.h> #include<conio.h> void main() { clrscr(); for (int i=1;i<=100;i++) { if(i%2 !=0 && i%3 !=0) { printf("\n Numbers that are not divisible by 2 or 3:%d \n ",i); } printf("\n"); } getch(); }
#include<stdio.h> #include<conio.h> void main() { int i,c,n; printf("Enter the number of terms you want which are divisible by 7"); scanf("%d",&n); c=0; for(i=7;;i+=7) { printf("\n %d ",i); c++; if(c==n) break; } getch(); }
Write a program which takes any number of days from the user. the program should display the number of years, number of months formed by these days as well as the remaining days.
write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? check bellow link http://bashscript.blogspot.com/2009/08/gcd-of-more-than-two-numbers.html
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
This isn't a question!
JAVA
To write a C++ program to display the student details using class and array of object.
$n = 10*(1+10)/2;
By learning how to program on C+.
(rand()&50+1)*2
Prime numbers are numbers that are only divisible by themselves and the number 1. You can write a program to print all prime numbers from 1 to 100 in FoxPro.
/* a number divisible by x should be a multiple of x */ int i, n,x; scanf ("%d", &n); scanf ("%d",&x); for(i=x;i =< n;i+=x) { printf ("%d", x); }
Identification division. Program-id. Quadratic. Environment division. Data division. Working-storage section. 01 a pic 9(3) value 0. 01 b pic 9(3) value 0. 01 c pic 9(3) value 0. 01 d pic 9(3) value 0. 01 e pic 9(3) value 0. 01 f pic 9(3) value 0. 01 g pic 9(3) value 0. 01 h pic 9(3) value 0. 01 x1 pic 9(3) value 0. 01 x pic z(3).z(2) value 0. 01 x2 pic 9(3) value 0. 01 y pic z(3).z(2) value 0. Procedure division. Display "Written by Martin O. Egua, but not complete". Display "Quadratic equation solver for three values a, b & c" Display "Enter a number: " Accept a. Display "Enter the second number: " Accept b. Display "Enter the last number: " Accept c. compute d = (b * b) compute e = 4 * a * c compute f = 2 * a compute g = d - e compute h = function sqrt (g). compute x1 = (-b) + h compute x = x1 / f Display "X = " x compute x2 = (-b) - h compute y = x2 / f Display "Y = " y Display "Send the accurate program". Stop run.
nr\m;laeoh9y0m g.qthnedxc In fortran: do i=1,5 write(6,*)i enddo stop end This program will write the numbers 1 to 5 on the screen.
you do this 10 print "0112358132134" use the whole of the thing