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For thousands of years, enlightened minds have struggled to decide which science is superior to all others and contains the knowledge needed to explain the world around us. It is our belief that Mathematics holds this prestige. Mathematicians are the ones who look for rules and algorithms behind an apparent chaos and infinite complexity.
Throughout time, humans have striven to discover the rules and patterns of the material world, to determine the qualities of objects, the complex relationships between them and what surrounds them. The math-game synergy has become a banality, the relationship between the two fields of manifestation of human intelligence being well exemplified in both senses. Despite being somewhat more difficult to illustrate, the transition from gaming to mathematics or vice versa is also well covered.

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The Game…. There have been thousands and thousands of games created for specific ages and for all ages... It may seem trite to talk about the educational-formative importance of games, about how they shape and train memory, analytical power, concentration, insight and flexibility in thinking, a capacity to evaluate complex situations or to anticipate.
Therefore, we should not be surprised by mathematicians interest in games, nor the interest in the mathematics of those attracted to logical games. It cannot be otherwise, many games include a quantity of mathematical thinking, and mathematics itself being, to a certain extent and in a certain meaning, a great “game against nature, against its unknown grammar”.

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Problem nr.1�

July and Stephen dealt their 52 cards on half. Cards from 2 to 10 are worth 10 points, the ace is worth 11 points, the jack 12 points, the queen 13 points and the king is worth 14 points. Stephen noticed that he didn’t have any aces in his stack and no 2’s cards, but he doesn’t have 4 equal worth cards, and July noticed that she has a maximum of 2 cards over 11 points that have equal worth.
July adds the values of the cards in her stack. What is the lowest value this amount can have? What about the biggest?
NOTE: A pack of playing cards contains 52 cards: 4 cards worth 2 points, 4 cards worth 3 points, 4 cards worth 4 points and so on until 4 cards worth 10 points, then 4 aces, 4 jacks, 4 queens and 4 kings.

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Solution(1)

July must have in her stack all of the 4 cards worth 2 points, the 4 aces and at least one of each card. This means :

4 + 4 + 11 = 19

19 cards known by July. We pick the other 7 cards so that their value is minimal: three more 3, three more 4 and one 5. We get the sum :

4*(2+11+3+4)+2*5+6+7+8+9+10+11+12+13+14=169 .

In this case, Stephen would have 2 cards worth 5 and 3 cards of 6, 7, 8, 9, 10, one jack, one queen and one king, a case in which we respect the conditions of the problem.

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Solution(2)

In order for the sum of the cards held by July to be maximum, the other 7 cards we don’t know must have as big as possible values. But July has no more than 2 jacks, 2 queens and 2 kings ( we already know she has one of each ), so, in order to get the maximum sum, we will assign her one more of each card. Because July already has 4 aces, she’ll have another 3 cards worth 10 points and one card worth 9 points. If we assign her with all 4 cards worth 9, then she’ll have 28 cards, which is not true because they must have only 26 cards each.
So, July has 4 cards of 2, 4 of 10 and 4 aces, 2 cards of each 9, jacks, queens and kings, plus one 3, 4, 5, 6, 7, 8. Stephen has 3 cards of each 3, 4, 5, 6, 7, 8 and 2 of 9, jacks, queens and kings. This way, the cards are split correctly.
Therefore, the maximal sum July can get is :

4*(2+10+11)+2*(9+12+13+14)+3+4+5+6+7+8=221.

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Problem nr.2

Simona and Ilie play “tension”: they toss a coin and if the result is “head”, then Simona gets a point, and if it is “tail”, then Ilie gets a point. Win the game against that player who has at least 4 points, with at least 2 points difference. The game is won by Simona with a score of 10-8. How many different tossing sequences lead to this result?

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Problem nr.3

Stefan, for his mail, is building his password by switching letters in his surname(STEFAN), coming up with the combination FANEST. Determine the number of switches.

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In our life, the majority of events which are occurring in the short or long term are random. This is how the necessity of studying the science of hazards has arrived. So it wouldn’t be wrong if we apply a little math from time to time, so we can increase the probability of knowing the final result beforehand.

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Problem nr.4

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Solution(1)

a) We call the two boxes: (i) and (ii). To reach the given situation, the person took out a box from his pocket (2n - k + 1) times. N times a box to empty it, (n - k) times the other box to have k sticks left, and again the first box to find that it is empty.
So the events take place:
A - "in 2n - k cases box (i) appears n times";
B- "in 2n - k cases box (ii) appears n times"
The two events obviously have equal probabilities. Let's consider the next events:

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Solution nr.2

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Problem nr.5

For the flooring of a room in the shape of an isosceles right triangle, with the length legs natural numbers, square tiles are brought which have the side length 1. They can be placed whole or cut into right triangles of legs 1. What is the minimum number of boards that must be cut to floor the hall?

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Solution

If the plates are placed as shown in the figure, then on each strip parallel to a leg are only used squares until the last position (next to the hypotenuse) where one triangle must be placed. As there are n strips, it follows that there are needed n half squares. We establish that no other floors are possible. If we start from the hypotenuse, putting in B a triangle and then only squares, we should have: k + 2 = n √ 2 (where k is the number of squares), so √ 2 = (k + 2) / n, impossible. If we start in A 'pushing' a square with two sides parallel to the hypotenuse, the remaining triangle is smaller than possible triangular plates. Other non-symmetrical attempts are excluded. It remains the only possible flooring. If n is even, then we cut out n / 2 squares, if n is odd, then (n + 1) / 2, which means N = [(n + 1) /2] squares.

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Bacteria and viruses are infectious entities that need a host organism to develop and multiply, this body gets sick most of the time. Bacteria are cells, viruses are not. The bacterial infection is treated with antibiotics, the viral one not, but with antivirals, a very complicated class of medication.

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Problem nr.6

A virus enters a colony made by n bacteria. In the first minute, it kills a bacteria, then it divides into 2 new viruses and simultaneously each of the remaining bacteria is also diving into 2 other ones. In the next minute, the 2 viruses kill 2 bacteria, then the 2 viruses and the remaining bacteria are diving again and so on.
Will the colony live forever or will it be killed after some time?

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