有三堆石子，每堆石子的總數分別為51顆、49顆及5顆。每次操作我們都可以選擇以下的兩種方式之一：(1) 若某堆石子的總數為偶數時，可以將此堆石子分為數量相等的兩堆；或(2) 將任意兩堆石子合併為一堆。試問：經過幾次會將上述的操作將這三堆石子分成105堆，且每堆只有一顆石子？
- 釋塵Lv 76 年前最佳解答
這題在環球城市數學競賽2001年春季賽國中組高級卷曾出過,因九章網站只收錄英文解析,小弟貼原英文解析,請參考! Since all three piles initially contain odd numbrs of stones, the first move must be a merge. If we merge the piles with 5 and 49 stones, then we have two piles in each of which the number of stones is a multiple of 3. We claim that from now on, the number of stones in any pile must be a multiple of 3. Since 1 is not a multiple of 3, not even 1 pile with 1 stone can be obtained. Note that the merger of two piles of stones each with a number of stones equal to a multiple of 3 will result in one pile with a number of stones equal to a multiple of 3. Splitting of a pile with an even number of stones does not change this property. Thus the claim is justified. Similarly, if we begin by merging the piles with 5 and 51 stones, then the number of stonesin every subsequent pile must be a multiple of 7. If we begin by merging the piles with 49 and 51 stones, then the number of stones in every subsequent pile must be a multiple of 5. It follows that we cannot get 105 piles each with 1 stone. 英文部份應該就說的很清楚了,希望有幫上你的忙!