Find the smallest number whose digits multiply to a given number n | GeeksforGeeks
Given a number ‘n’, find the smallest number ‘p’ such that if we multiply all digits of ‘p’, we get ‘n’. The result ‘p’ should have minimum two digits.
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Given a number ‘n’, find the smallest number ‘p’ such that if we multiply all digits of ‘p’, we get ‘n’. The result ‘p’ should have minimum two digits.
Case 1: n < 10 When n is smaller than n, the output is always n+10. For example for n = 7, output is 17. For n = 9, output is 19.
Case 2: n >= 10 Find all factors of n which are between 2 and 9 (both inclusive). The idea is to start searching from 9 so that the number of digits in result are minimized. For example 9 is preferred over 33 and 8 is preferred over 24.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.
void findSmallest(int n){ int i, j=0; int res[MAX]; // To sore digits of result in reverse order // Case 1: If number is smaller than 10 if (n < 10) { printf("%d", n+10); return; } // Case 2: Start with 9 and try every possible digit for (i=9; i>1; i--) { // If current digit divides n, then store all // occurrences of current digit in res while (n%i == 0) { n = n/i; res[j] = i; j++; } } // If n could not be broken in form of digits (prime factors of n // are greater than 9) if (n > 10) { printf("Not possible"); return; } // Print the result array in reverse order for (i=j-1; i>=0; i--) printf("%d", res[i]);}