The sum of the first 40 positive integers (1-40) is: 820
226
#include <stdio.h> #include <conio.h> void main() { int d[3][3] = { 1, 2, 6, 3, 8, 5, 5, 6, 7 }; int k = 0, j = 0; int sum1 = 0, sum2 = 0; for (j = 0; j < 3; j++) { for (k = 0; k < 3; k++) printf(" %3d", d[j][k]); printf("\n"); } for (j = 0; j < 3; j++) { sum1 = sum1 + d[j][j]; } k = 3 - 1; for (j = 0; j < 3; j++) { if (k >= 0) { sum2 = sum2 + d[j][k]; k--; } } printf("Sum of First diagonal= %d\n", sum1); printf("Sum of Second diagonal= %d", sum2); getch();
/* @Autor: MD moniruzzaman http://www.youngprogrammer.com */ #include<stdio.h> #define maxn 5 int matrix[maxn][maxn] = { {1,2,3,3,4},{2,3,4,1,2},{ 4,5,6,7,8},{3,4,5,6,9},{4,3,2,1,0}}; /* Given matrix is: 1 2 3 3 4 2 3 4 1 2 4 5 6 7 9 3 4 5 6 9 4 3 2 1 0 */ int main() { int sum = 0, i, j; for(i = 0; i<5; i++) { for(j = 0; j<5; j++) { sum+= matrix[i][j]; } } printf("%d\n",sum); return 0; }
Straight line depreciation method allocate equal amount for all years while in sum of years digit method depreciation is allocated with high amount in initial years while low amount in later years.
They are: 99+67 = 166 and 99-67 = 32
The sum of the first ten positive integers is: 55
The sum of the first 500 positive integers is: 125,250
The sum of the first 30 positive integers is: 465.
The sum of the first 60 positive integers is 1830.
The sum of the first 200 positive integers is 19,900.
The sum of the first eleven positive odd integers is 121.
The sum of the first 2,006 positive, odd integers is 4,024,036.
The first odd positive integers are "1" and "3" which the sum is 4.
The sum of the first thousand even, positive integers is 1,001,000.
The sum of the first 40 even positive integers can be equal to 820.
The sum of the first 5,000 odd, positive integers is 5,000 squared or 25,000,000.
The sum of the first 230 positive, odd integers is 230 squared or 52,900.