### Solved Problem 21

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21/README 0 → 100644
 Question: Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220. Evaluate the sum of all the amicable numbers under 10000. Answer: 31626
21/main.cpp 0 → 100644
 #include #include using namespace std; int main() { bool found = false; int sum = 1; int max = 10000; vector numbers; vector final_numbers; numbers.push_back(0); numbers.push_back(1); numbers.push_back(3); numbers.push_back(4); for(int i = 4; i <= max; i++) { sum = 0; for(int j = 1; j < i/2; j++) { if(i % j == 0) { if(j == (i/j)) { sum += j; break; } if(j > i/j) break; if((i/j) != i) sum += i/j; sum += j; } } numbers.push_back(sum); } sum = 0; for(int i = 4; i <= max; i++) { if(numbers[i] <= max && numbers[numbers[i]] == i && i != numbers[i]) { found = false; for(int j = 0; j < final_numbers.size() && !found; j++) if(final_numbers[j] == i) found = true; if(!found) { final_numbers.push_back(i); final_numbers.push_back(numbers[i]); sum += i + numbers[i]; } } } cout << sum << endl; return 0; }
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