Divide some peanuts into three groups and one peanut is outstanding. Mix two groups thereof with the outstanding peanut and further divide them into three groups. Then two peanuts are outstanding. Mix group thereof with two outstanding peanuts and further divide them into three groups. Then they are no more peanuts outstanding. How many peanuts are there at least?
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To solve this problem, let's break it down step by step:
Step 1: Divide some peanuts into three groups and one peanut is outstanding.
Let's assume the total number of peanuts is X. After dividing X peanuts into three groups, we can represent the number of peanuts in each group as (X/3, X/3, X/3), where the last peanut is outstanding.
Step 2: Mix two groups thereof with the outstanding peanut and further divide them into three groups. Then two peanuts are outstanding.
From step 1, we had three groups of (X/3, X/3, X/3), and we mixed two of these groups with the outstanding peanut. This gives us (X/3, X/3, X/3, 1). Now, we divide them into three groups, which would result in (X/9, X/9, X/9, 1/3, 1/3, 1/3).
Step 3: Mix group thereof with two outstanding peanuts and further divide them into three groups. Then they are no more peanuts outstanding.
From step 2, we had six groups of (X/9, X/9, X/9, 1/3, 1/3, 1/3), and we mixed one of these groups with the two outstanding peanuts, resulting in (X/9, X/9, X/9, 1/3, 1/3, 1/3, 2/3, 2/3). Now, we divide them into three groups, which would give us (X/27, X/27, X/27, 1/9, 1/9, 1/9, 2/27, 2/27, 2/27).
Step 4: No more peanuts outstanding.
From step 3, we can see that there are no more outstanding peanuts. Thus, the minimum number of peanuts is X/27.
So, the minimum number of peanuts is X/27, where X represents the total number of peanuts at the beginning.