Question Number 179574 by Ar Brandon last updated on 30/Oct/22
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\:\mathrm{less}\:\mathrm{than}\:\mathrm{400}\:\mathrm{and}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{6}. \\ $$
Commented by Ar Brandon last updated on 30/Oct/22
#include <stdio.h> int main(void) { int sum = 0; for (short i = 1; i < 400; i++) if (i % 6 != 0) sum += i; printf("%d", sum); return 0; }
Commented by Ar Brandon last updated on 30/Oct/22
Output: 66534 Any mathematical solution?
Commented by mr W last updated on 31/Oct/22
$${Ar}\:{Brandon}\:{sir}: \\ $$$${can}\:{you}\:{please}\:{calculate}\:{the}\:{answer}\:{to} \\ $$$${Q}\mathrm{179420}\:{with}\:{a}\:{program}? \\ $$$${it}\:{is}\:{total}\:{number}\:{of}\:{numbers}\:{between} \\ $$$$\mathrm{10}\:{and}\:\:\mathrm{9876543211}\:{which}\:{are}\:{divisible} \\ $$$${by}\:\mathrm{9}\:{and}\:{have}\:{different}\:{digits}. \\ $$$${the}\:{first}\:{one}\:{is}\:\mathrm{18}\:{and}\:{the}\:{last}\:{one}\:{is} \\ $$$$\mathrm{9876543210}. \\ $$
Commented by Ar Brandon last updated on 31/Oct/22
OK Sir, I'll try it this evening. We are going to start from 18 and increment by 9 then store the remainders when divided by 10, 100, 1000, etc in arrays then use the nested for loop to compare each value stored in the arrays by the others. If the values are all different they pass the test, else they're ignored. See you in the evening, Sir ���� Thanks for mentioning me for this task.
Commented by Ar Brandon last updated on 05/Nov/22
#include <stdbool.h> #include <stdio.h> int main(void) { bool flag; int num[10]; long int count = 0; for (long int i = 18; i < 9876543211; i += 9) { num[0] = -1; num[1] = -2; num[2] = -3; num[3] = -4; num[4] = -5; num[5] = -6; num[6] = -7; num[7] = -8; num[8] = -9; num[9] = -10; num[0] = i % 10; num[1] = i % 100 / 10; if (i > 99) num[2] = i % 1000 / 100; else goto compare; if (i > 999) num[3] = i % 10000 / 1000; else goto compare; if (i > 9999) num[4] = i % 100000 / 10000; else goto compare; if (i > 99999) num[5] = i % 1000000 / 100000; else goto compare; if (i > 999999) num[6] = i % 10000000 / 1000000; else goto compare; if (i > 9999999) num[7] = i % 100000000 / 10000000; else goto compare; if (i > 99999999) num[8] = i % 1000000000 / 100000000; else goto compare; if (i > 999999999) num[9] = i / 1000000000; compare: for (short j = 0; j < 10; j++) { for (short k = j+1; k < 10; k++) { flag = true; if (num[j] == num[k]) { flag = false; goto loop_end; } } } loop_end: if (flag) count++; } printf("%ld\n", count); return 0; }
Commented by Ar Brandon last updated on 05/Nov/22
I tried this, Sir! But got no output. It seems the figures are too large for my 32 bit compiler. Or perhaps I've missed something.
Answered by manxsol last updated on 30/Oct/22
$$\sum_{\mathrm{1}} ^{\mathrm{399}} −\mathrm{6}\sum_{\mathrm{1}} ^{\mathrm{66}} \:\:\:\:\:\:\:\:\:\sum_{\mathrm{1}} ^{{n}} ={n}\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$$${integer}\left(\frac{\mathrm{400}}{\mathrm{6}}\right)=\mathrm{66} \\ $$$$\mathrm{399}×\frac{\mathrm{400}}{\mathrm{2}}−\mathrm{6}×\mathrm{66}×\frac{\mathrm{67}}{\mathrm{2}}=\mathrm{66534} \\ $$