I enjoyed walking 
In my opinion, since each one separately would have spent on your pair of shoes a different amount, should share the money they have saved in proportion to what they were going to spend. There may be people who think the savings to be divided equally, or otherwise, but I think it should be in proportion to the money they intended to spend, because the product that takes one is more valuable than that of another, and must pay more for it. Once
made it clear the objective, we do accounts: Carmen has been a product of 2000 while George has been one of 3500, meaning that for every 20 euros of value two carmen, Jorge takes 35 or a ratio of 4 to 7.
That means that for every 11 euros to pay, should pay 7 Jorge and Carmen 4. Since they have been paid between the two 3500, by dividing this number by 11, gives 318 with 18 cents (there is no exact), in my humble opinion, Carmen would have to pay 318.18 * 4 = 1272.73 (rounded) and Jorge 318 , 18 * 7 = 2227.27, making a total of 3500. Looking
what you have saved, Jorge will have saved 1272.73 and 727.27 Carmen, which in both cases corresponds to a savings rate of 36.36%.
To attack this problem well there are several approaches, but I have been very useful by the fact that the centroid is exactly twice as far from the vertex to the midpoint of the opposite side.
This fact is also used to show that the three medians intersect at a single point.
The idea is to draw the segments perpendicular to the line, not only from the vertices, if not from the midpoints of the sides.
line and perpendicular Baricentro
In our picture, the vertex that is separated from the other by the line I call A, and two B and C. Obviously, A 1, which is the midpoint of segment AB is on the same side as B and C and B and C 1 1 are on the same side as A, because if the line could not pass the cutoff.
Now, by similarity, we have the segment from C is twice that of C 1 , that of B is twice that of B 1 and A is twice that of 1 .
As the midpoints of the segments divided into two equal parts on the sides, the segments defined with respect to our line also play a special relationship. Indeed, it is easy to see that the segment unless the segment of B 1 is exactly like the B segment 1 More segment C. Likewise, the segment of B minus 1 A segment worth the same as the segment of A 1 less than C, and the segment of A less than C 1 worth the same as the of B plus C 1 .
Let us now try to show the equality that we demand. We start from the sum of the segments B and C. As we have seen that the segment of B plus C 1 is equal to A less than B 1 , we need the segment of C is equal to the segment unless the segment twice 1 B . Therefore, the sum of the segments B and C is equal to the sum of B plus A least twice the segment B 1 . But we also know that the segment of B is exactly twice that of B 1 therefore conclude that the sum of the segments B and C matches that of A.
To the first question, actually, is to buy time, so the regularity view, we accept the suggestion made by Miguel Escobar in the comments, add 1 + 2 + .. + 8 and 2 + 3 + ... + 9 separately, using the "trick" of the matches, taking (1 + 8) * 4 and (2 + 9) * 4, ie 36 and 44. As the scores are multiplied by 10, add 360 + 44 = 404, which is the correct solution (B). For
the second question, obviously the number of divisions must be a common multiple, and the youngest is 30 (D).
The third question is to think a little more, is a type of puzzle sudoku. The 9 can only add 11 with a single number, so it must go in one end and 2 at the intersection.
for 8 Now we have two possible positions. Suppose we place it in the next round, we force the intersection to be 1 and we quickly came to a contradictory situation.
Then the other end is occupied by the 8 and the next intersection by 3.
A quick score leaves us companion of 2 to 5, which requires the 4 at the next intersection, to 7 on the other side and at the intersection and the place that interests us in this setting is 6.
Can there be other configurations (in addition to symmetric)? The answer is no, but in this type of test should not stop to check. When we find a valid configuration, you must respond. And the answer is D (6).
Actually, there is an argument even faster. Once we have placed the 9 and 8 at the ends, the two central circles added 22, the ends 8 and 9 are the sum total to 39, and as the sum of the numbers 1 through 9 is 45, the central should be, as we know, 6.
As every midday Lucia gave a smile to hear the second chime to the echo alone naughty always added to the ringing of the church tower. Of course, there were still two, so it was good time to get to the restaurant, and continued putting on makeup smiling as she wondered why this phenomenon is repeated every day I still find it funny. But that day, perhaps because they were preparing to celebrate an anniversary, he gave him time to think about some phenomena that need to happen, in how long it takes sound to travel through the valley, or how much you need a ray of light to get from a star. Then let the lipstick and was looking into her eyes, to her reflection. Frozen. Pensive. 
These games should always start at the end, to consider how you can ensure victory, and dividing the situations in which values \u200b\u200band values \u200b\u200bto win you lose.
In our case, we must begin the study from 100, which is the first time that someone wins. Obviously, 93 and 96 are positions from which we can win, we must prevent our opponent has them. Looking for a number with which to force our opponent to return that value, are suddenly 89. If our opponent adds 4, it outputs 93, and if you add 7, 96. In either of both, we win. So really, whoever gets to leave by 89 wins.
Reasoning backward, find the 78, 67, 56, 45, 34, 23, 12 and 1. If we can pass 1 to our rival, sure we would get to 100 and we would win.
Unfortunately, we must begin by adding 4 or 7. So we should not point to 100. If we repeat the system (note that is to subtract 11 each), we reason from 200 we should start from 2, to reach 300, from 3 to at last!, To get to 400, from 4.
Now let's check if our system works. We started by adding 4. Our opponent makes his move, and we do the opposite, so that they reach 15. We will continue, adding the move contrary to what our opponents do. When we get close to 100, we pass the rival 4 + 11 * 8 = 92, so that you can not get 100, if not 96 or 99. We add the number opposite, reaching 103, and proceed. The next step will be dangerous 103 + 8 * 11 = 191, in which our opponents can get to 195 or 198. Join us again to reach 202. The final step will be to dangerous 202 + 8 * 11 = 290, in which our opponents can go to 294 or 297. We arrive in any case at 301. The end of the game will arrive in the 301 + 8 * 11 = 389, which we pass to rival that number, so that brings us back 393 or 396, winning in any case to get 400.
To find these numbers wisely, we must see what we can deduce general characteristics. The only thing I can think is that the first number can only be 1, 2 or 3. Once
view this restriction, you have to start scoring. The best, in this case, is to start with the last digit of the first issue, since we can quickly calculate the last digit of the other two (and note whether or not we get and how much). After the penultimate digit, keeping an eye that does not generate repetitions, and then the first digit. Obviously, if generating the last or the second digit "spend" on 1, 2 and 3, we find no possible value for the former, as we said before.
Thus it is clear that the last digit can not be 1. If you look at 2, we find after some trial and error numbers 192, 384 and 576, which meet the statement.
If we exhaust all possibilities, we can continue testing for other possible last digits. By 3, we found 273, 546 and 819. The 4 does not provide any trio after playing a while. The 5 is immediately discarded. The probe 6 has cost me a lot and not find anything.
However, with 7 meeting 327, 654 and 981, and 257, 534 and 801. With 8 not get any fruit and the last I get to 9: 219, 438 and 657. In total five different possibilities meeting.
In this issue, it's best to try to try to build you an example. Obviously, if we have to divide many times, it is best to try 1, which is the universal divisor, and the first three candidates that tends to happen is 1, 1, 1, which clearly, each number divides the sum of the others.
already have one, we can now begin to work more. Since there can be other cases relating to same (it would not be a primitive triplet), what happens if there are two equal and the other is no longer the same? That is, if we have a triple (n, n, m) or (m, n, n) with this property, we have that m divides 2n. If different n, or exactly 2n, in which case we reach the triple (1, 1, 2), and they can not have common divisors, or is less than n. In the latter case, once you take away all common divisors with n, can only be 2 or 1. However, if 1, this requires that n must divide n + 1, which is impossible for n greater than 1, or n must divide n + 2, which can not be for n greater than 2.
Well, we have already removed the case question of two or three equal numbers, we will deal with different numbers.
Suppose the lowest is 1. The second, ie, b, c should be divided, but the third, c, must divide ab + 1, so it must be b + 1 (otherwise you would not be greater than that by less than or equal to b + 1) . However, b must divide ab + 2, so it must divide 2, and that means it's 2 (can not be 1, as it would like to). Therefore, we have another short list: (1, 2, 3).
If the child is not 1, there is a prime p which divides it. Working with cousins \u200b\u200bis very comfortable as you can keep track on a par. That cousin can not divide another issue boc, because if so, divides the three. This is proved as follows: as to divide divides ab + cyp a, p divides ab + c, so who can divide only one of them.
Now for the greatest number. Since it is greater than b, and divide a + b, must be less than a + b, and greater than b. As the largest factor of a + b (after himself) is less than b, since a + b is less than 2b, c must match a + b. So our slate is of the form a, b and a + b. However, b must divide 2a + b, so you must divide 2a. That is reduced, since it has no common factors with a, b is 2 to 1, cases have been treated. So
the only valid ordered triples are (1, 1, 1), (1, 1, 2) and (1, 2, 3).
This is a very hard for this age group, but in fact all you need is to realize a series of relationships between numbers.
To fix ideas, we try to find some values \u200b\u200bfor the function in case we are asked (r = 5 s = 8), to predict what value F (2010), as requested in paragraph (1). Let's try
F (1). As F (1) = 1 + 5 or F (1) = 1 to 8, since it is a positive integer, the latter possibility has been discarded then F (1) = 6. Likewise, F (2) = 7, F (3) = 8 and so on, until we get F (9), in which we can choose between 9 + 5 and 9 to 8, because both are positive . However, if you think about it, we find that there is no alternative to choose from 9 to 8 = 1, since all numbers should be F (n) for some n, and 1 must be, therefore, of the form n - 8 or of the form n + 5. Of course, that 1 can not be of the form n + 5, so it must be F (9). Thus 1 = F (9), 2 = F (10), and so 5 = F (13).
words, the first 13 values \u200b\u200bbecome, successively, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4 and 5. Now something similar happens again: F (14) can not be 14 - 8 = 6 because 6 is "used" (that is F (1)), then F (14) = 14 + 5 = 19. In summary, the following values \u200b\u200bare 19, 20, 21, 22, 23, 24. 25, 26, and then return to use the expression n - 8, because if we do not, the numbers 14 to 18 shall be without being the image of any number. So we continue with 14, 15, 16, 17, 18.
guess which clearly shows what the pattern is not it? Actually use the set of positive integers of 13 in 13 using the formula n + 5 in the first 8 and the formula n - 8 and the other five. Let's see what happens with the 2010. The
2010, if we try to divide by 13, gives 154, and a remainder of 8, ie 2010 is the eighth in the group that begins after 2002 = 154 * 13. After you apply the formula n + 5 and hence F (2010) = 2015.
Let's try now to answer the second part. In general, if r and s are not worth 5 and 8, if no other quantities to calculate F system is similar. Applies to sets of consecutive numbers r + s, so that the s first formula is applied to n + r, r and the following formula is applied n - s. It is very easy to extend the reasoning that we used before the general case.
To understand the second part, also applied to the case r = 5, s = 8, and then try to apply it in general. Since each set of 13 consecutive numbers become themselves the same way, simply look at the 13 prime numbers, because if you meet them, will be true for any set of numbers. Apply
F twice done, for example, that F (F (1)) = F (6) = 11. Obviously not enough. Applied 3 times, makes F (F (F (1))) = F (11) = 3. If we apply it four times, we will have 4 F (1) = F (3) = 8. If we apply again F 5 (1) = F (8) = 13. Thus, we will successively by 5, 10, 2, 7, 12, 4, 9, and finally arrive 1 in F 13 (1) = F (9) = 1. With a little work, we find that 13 is the lowest of times you have to apply this function to reach 13 F (n) = n.
Will the sum of r + s the answer? No, but before reaching it, see what happens when we apply repeatedly F. In fact, add subtract r s repeatedly, trying not to get out of the set of 1 to r + s, so we always results as ar - bs, where a and b are positive numbers. Combinations of integers are well known to fans of math problems, since you can always be achieved by choosing correctly and b, giving lower value as a result equal to the greatest common divisor of r and s.
That is, the number nearest to 1 on the pass will be 1 + d, where d is the greatest common divisor (in the case of 5, 8, passed by 2, since 1 = 5 * 5 - 3 * 8). Of course, we also 1 + 2d (3 in our case) and all terms of an arithmetic progression of difference d, ie, d hopping tours all numbers r + s. In the present case, since 5 and 8 have a GCD of 1, we have to give 13 jumps back to the original. Moreover, as we can not agree to the same value in two different dives, we will give exactly 13 jumps starting from any number.
Now if rys have another MCD, then it is impossible to achieve with subtraction of sums of r s values \u200b\u200bless than d = GCD (r, s), so just go (r + s) / d values, ie that the answer is actually the second paragraph is worth k (r + s) / MCD (r, s). For example, if r = 4 s = 6, k stands 5 (path 1 would be: 5, 9, 3, 7, 1).
Statement This time, again, the comments I have removed much of the work.
The first question is simple. Indeed, as Matthew has to climb 12 stories, half of road has risen 6, if in the 8 th game is that it has 2, so that Clara must live in the 14 th, that is the answer (c).
The second response, M = 1000 and CD = 400 (since C is 100 and D 400). Since XL = 40 (similar), IX = 9. So we have to MCDXLIX = 1449. Corresponding to the response (a).
Solve the third question is more complicated. In reality, we need to find out what it's worth every letter, and as we have the products of each pair is for me and try to factor it easier to multiply and divide quickly head. Thus, xy = 2 * 3 * 3, xz = 3 and z = 3 * 2. If we multiply the second by the third equality, we have x * y * z * z = 3 * 3 * 2, and if we divide by the value of x * and also have it, is that z * z = 1, ie z = 1, being positive . Quickly obtain that x is 3 and y is 3 * 2 = 6. Ie x + y + z = 10, response (b).
Another approach (faster) could be that since xz = 3 yxyz are positive integers, or x = 3 and z = 1, where y = 6 (and the sum is what we have said), or x = 1 and z = 3, but in this case, and should be worth 2, and could not assert x 18. Therefore, we only have the solution (b).
Statement several ways to solve these problems without resorting to formal algebra. The idea is to play with watermelons and melons, giving values \u200b\u200bto their weights, putting more together, and so on.
A nice way to solve is this: if three and four melons, watermelons weigh 13 and four watermelons and three cantaloupes weigh 15, to put it all together we have seven and seven melons watermelons weigh 28. Now, if we make seven parts, each couple have a watermelon and a cantaloupe weigh 4.
Having this idea in mind, we return to a track in the beginning. As three and four melons watermelons weigh 13, and is very close to being matched quantities, try to reach a similar figure.
now know that three watermelons and three cantaloupes are three couples as we have done before, which in total weighed 4 + 4 + 4 = 12. That means that the melon is most in the initial track weighs 13 to 12 = 1kg, so that each watermelon weighs 3kg.
We found that both tracks are true, ie three and four melons, watermelons weigh 9 + 4 = 13 and four watermelons and three cantaloupes weigh 12 + 3 = 15.
did not know her too. Enough to know that nothing in life was easy, on certain afternoons walked to the lighthouse with his childhood friend and sometimes he was accompanied by a cold coffee on the terrace of the port. No one remembers if fortune ever walked at his side. What if one day the chance was wrong, and we were graced by the lottery all, she had lost her ticket. That "s" the dictionary was probably the only place you ever found luck. 
The idea is that there can be equal to a proper fraction, ie it is not a regular number.
To demonstrate this statement, we will support the demonstration of the irrationality of √ 2, which normally must all have seen in class.
In case you have not seen I do a quick check. We proceed by contradiction, assume that the value of this number is a fraction a / b, a and b coprime, because if they were not, simplify the fraction. After taking equality and denominator roots are removed and the expression is reached 2b 2 = a 2 , so that 2 is even. Because of that, a must be even, as if it were odd, 2 should also be odd. Then a = 2k for some k, and substituting in the previous equality, we come to 2b 2 = 4k 2, so 2 b 2 = 2k, where we also b is even, which is absurd, since we had assumed that the fraction a / b could not be further simplified. Hence, √ 2 can not be rational.
applied to the problem at hand, we can start proving that for all natural n, √ n must be either integer or irrational. As
proceed by reductio ad absurdum. If not an integer or irrational, √ \u200b\u200bn is of the form a / b, and we can simplify the fraction to a and b are relatively prime. Also, b must be greater than 1, because if not, √ n would be over.
Now suppose that p is a prime that divides n (n = pk). If we remove the root squaring and denominators we arrive at the following equation with integers, nb = PKB 2 2 = a 2 , then p divides 2 aa, and therefore must divide a. That means that a = pt, and the equality becomes PKB 2 = p t 2 2 . Simplifying kb 2 = pt 2 . As p does not divide ab, he was a cousin, you must divide k, so k = ps, so 2 sb = t 2 , and we can repeat reasoning. That means that all prime factors of p are repeated twice, which means that its root is full. Which was discarded from the outset.
can also be reasonable to assume that we have found the smallest n for which this expression is given, and the method we have built, you come to another value s less than n is of this form.
Once seen, we can build another similar reasoning to cube roots, as the method to remove a cube root is simply to raise the bucket, and would get another expression of the form nb 3 = PKB 3 = to 3 and reasoner in the same way. In fact, it can be shown for roots of generic index.
Now imagine that we have natural nym. If the expression √ n + √ m 3 = a / b, a and b mutually prime and b> 1, then √ n can not be whole because we could clear 3 √ m in the form of an irreducible fraction, we know that is impossible. Nor 3 √ m can be whole for the same reason, ie, we could clear √ n would be a fraction. However, equality √ n + √ 3 m = a / b we can remove and clear denominators, leaving b * 3 √ m = a - b √ n. Raise the bucket to remove the cube root, and get 3 mb = a 3 - 3rd 2 √ n + b 3anb 2 - 3 nb * √ n (this can be checked if we develop cube subtraction, multiplying a polynomial or three times a - b √ n and applying properties of the square root). Grouping
roots, we have the expression 3 mb = a + 3anb 3 2 - (3a 2 b + nb 3) √ n, so we can solve for √ n, becoming equal to 3 + 3anb 2 - 3 mb = (3a 2 b + nb 3) √ √ n and thus n = (a + 3anb 3 2 - 3 mb) / (the 3rd 2 b + nb 3). Thus, √ n should be rational or integer, which can not by reasoning that we have done before. Then the sum
√ n + √ m 3 is either an integer, or is an irrational number, we wanted to prove.
