# Mock AIME 5 Pre 2005 Problems/Problem 2

We have . If , then , but , a contradiction. If and , then , a contradiction. If and , then , a contradiction. If and , then , a contradiction. If and , then , a contradiction. If and , then , a contradiction. If and , then , a contradiction. If and , then and . Thus, we must have cde=2(hij), where are distinct digits from the list . If , then we have , a contradiction. Thus, we must have , and therefore . If , then we have , so . If we have then , a contradiction. If we have then (as is not in the list of permitted digits). Thus, we must have . If we have , then , a contradiction. If we have , then , which is not in the list of permitted digits, a contradiction. If we have , then , a contradiction. Thus, we must have , and therefore . But now we must have odd as . Thus, we have and . Thus, our minimal responsible pair of two 5-digit numbers is abcde=26970, fghij=13485. So, we have b+c+d+i+j=6+9+7+8+5=35. ~AbbyWong