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 the peanuts into three groups and have one outstanding peanut.
Let's start by assuming there are "x" peanuts initially. So when we divide them into three groups, each group will have x/3 peanuts, and we have one outstanding peanut left.
Step 2: Mix two groups with the outstanding peanut and further divide them into three groups.
Now we take two of the groups (each with x/3 peanuts) and mix them with the outstanding peanut. So, we have a total of (x/3) + (x/3) + 1 = (2x/3) + 1 peanuts.
Next, we divide these (2x/3) + 1 peanuts into three equal groups. Each group will have ((2x/3) + 1)/3 peanuts.
Step 3: Two peanuts are outstanding.
After dividing into three groups, we are left with two outstanding peanuts. So, we have ((2x/3) + 1)/3 = 2 peanuts.
To find the minimum number of peanuts, we solve this equation:
((2x/3) + 1)/3 = 2
First, let's simplify the equation by multiplying both sides by 3:
2x/3 + 1 = 6
Now, let's isolate x by subtracting 1 from both sides:
2x/3 = 5
Multiply both sides by 3/2 to solve for x:
x = 15/2 = 7.5
Since we can't have half a peanut, we can conclude that there must be at least 8 peanuts.
So, the minimum number of peanuts required is 8.