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In three farms there are a total of 333 animals. In the first farm there are triple the number of animals than in the second and in the second, twice as much as in the third.

**How many animals will have to be passed from the first farm to the others so that the number of animals in each one is a different three-figure number?**

#### Solution

We know that in the first farm there are three times as many animals as in the second and in the second, twice as much as in the third. That means that for every animal in the third, in the second there are two and in the first there are six. In this way, we have placed a total of 9 animals on the farms. If we place them 9 in 9, respecting this proportion, we will obtain 333/9 = 37 animals in the third, 74 in the second and 222 in the first, which is the amount that respects the conditions.

Now we have to get, passing animals from the first to the second and third three different Capicúas amounts, but that add 333. We need three Capicú numbers of three and different figures, that add 333. Obviously, in the three we will start and end with 1 , but for them to be different, the central numbers must add up to 3, so they will be 0, 1 and 2. That is, the quantities will be 101, 111 and 121. The simplest would be **pass 101 - 37 = 64 to the third, and 111 - 74 = 37 to the second**, leaving 222 - (64 + 37) = 222 - 101 = 121 in the first. But there are other solutions, which leave the three capicúas in different farms.

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Wonderful, it is a precious answer

I hope that the second part will be no worse than the first

I agree with told all above. Let's discuss this question. Here or in PM.