Mr. Euritis wants to distribute 2008 euros between Alberto, Benito and Carlos as follows: give € 1 to Alberto, € 2 to Benito, € 3 to Carlos, then € 4 to Alberto, € 5 to Benito, € 6 to Carlos , and so on, until you run out of money or cannot deliver the amount you touch.

**How much money does Benito receive and how much money will Mr. Euritis have left?**

Extracted from the problemate.blogspot.com.es page.

#### Solution

If we try to calculate how many laps Mr. Euritis takes, adding one by one the euros he gives, we will probably get tired, but there is a trick.

For example, in the fourth round, we will have to add 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12, which is a long sum. If we group it in a certain way, it probably looks like something else. You will see that if we put 1 + 12 + 2 + 11 + 3 + 10 + 4 + 9 + 5 + 8 + 6 + 7, there is an interesting regularity: every two addends add the same, that is, it is 13 + 13 + 13 + 13 + 13 + 13, or what is the same 13 six times. And it's six times because it's half of the total sum. Since 13 * 6 = 65, we are still far from 2008, but we have a method.

What if the number of addends is odd? Well, it also comes out. Imagine that we add up to the fifth round, we have to 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 1 + 15 + 2 + 14 + 3 + 13 + 4 + 12 + 5 + 11 + 6 + 10 + 7 + 9 + 8 = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 8 = 16 * 7.5 = 120. Note that although we match all the In addition, one is left without a partner, since it is odd, but it is worth exactly half the sum of two, so it is as if you had the exact half of addends than before.

If we assume that he has done n laps adding euros, we will have distributed 1 + 2 + 3 + ... + 3n = (3n + 1) * 3n / 2, that is, the sum of the last plus the first for the total of addends match For 2.

And this number we want to approach 2008, that is, (3n + 1) * 3n / 2 = 2008 (approximately). If we solve it as if it were an equation, we multiply by 2, and it remains (3n + 1) * 3n = 4016, where 9n2 + 3n = 4016, and we have the second degree equation 9n2 + 3n - 4016 = 0. Sure , that we are not looking for the exact solution, but the one that closest to 2008 produces the result without going over. The negative solution does not interest us, so we would have to n, approximately, (-3 + √ (9 + 4 * 9 * 4016)) / (2 * 9). The expression inside the root is 144585, whose root, rounding, is 380. If we subtract three it is 377, which, dividing it by 18 does not give exact, we round down to 20.

For n = 20, the sum gives 1830, while for 21 gives 2016.

Then in the money delivery series, it will stop at number 20, and **will be left over 2008 - 1830 = € 178**.

How much have you given to Benito? To calculate this, we must add 2 + 5 + 8 +… + 56 + 59, which will be the sum that we place in position 20. Again, we observe that 2 + 59 = 5 + 56 = 61, if we choose the 20 addends of convenient way we will have 10 times 61, which means **a total for Benito of € 610**.