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PROVA COMPLETA OBMEP 2026 NÍVEL 2 SEGUNDA FASE GABARITO E SOLUÇÃO DE TODAS AS QUESTÕES — Transcript

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0:00Hello everyone, let's solve the level
0:02two exam from the second phase of the
0:052026 MAP. Julia uses colored letters in
0:08place of the digits zero through nine
0:11to write equations. Different letters
0:14represent different digits. Item A.
0:17Initially, Julia writes the equations
0:20below. B + B = B. F x F = F. Which
0:24digits do the letters B and F represent
0:28? Since we already know that Julia
0:32chose each letter to represent a
0:34single-digit number, we need to see
0:36which of the digits from zero to nine
0:39satisfies the property that when I add
0:41it to itself, the result does not
0:43change. And what is the number such
0:46that? When I multiply it by itself, the
0:49result does not change. An element like
0:52this is what we call a neutral element.
0:56And here in this first expression, we
0:58need to find out what the neutral
0:59element of addition is. So imagine that
1:02, for example, this B here was one. If
1:05B were 1, it would mean 1 + 1 equals 2.
1:09That would already give a different
1:10result. If I put 2 + 2 it would be 4.
1:13So, if I add any number that is not
1:16zero, the result will give a value that
1:18is different from itself. Thus, the
1:21only option for this equality to be
1:24true is for B to be zero, because 0 + 0
1:27corresponds to zero. So, we have just
1:31discovered the value that the letter B
1:33represents. It is zero. Now for the
1:36letter F, adopting the same strategy,
1:38if I thought about putting 1 x 1, it
1:40would give exactly 1. Which is
1:44different. And 3 x 3 is 9. Any value
1:47greater than three will give a
1:49two-digit number that I cannot put here
1:51. So this means that the value of F
1:54must be one because 1 x 1 corresponds
1:57to 1. Therefore, we discovered that the
2:00letter F is worth 1. Thus, the digits
2:03representing the letters B and F are
2:04exactly the neutral elements of
2:06addition, which is zero, and the
2:08neutral element of multiplication,
2:10which is one. Item B. Afterward, Julia
2:13writes the equations below. J = J x J x
2:19J, which is equal to A. Which digits do
2:22the letters J, C, and A represent? We
2:26already know that zero and one are out
2:28of the game, since they represent the
2:30letters B and F, as we saw in the
2:31previous items. And now we have an
2:35exponentiation. What does an
2:37exponentiation mean? When we write a
2:41number raised to the power of another,
2:42placing the other number up here in
2:44small print, we call this bottom number
2:46the base and the number up here the
2:48exponent. This nomenclature means that
2:53I will take the bottom number and
2:55multiply it by itself a total of times
2:58equal to the number up here. So imagine
3:01, for example, that I were writing 2
3:04raised to the power of 4. When I write
3:07something like this, the two down here
3:09and the small four up here, it means I
3:12will multiply two by itself a total of
3:14four times. Thus, 2 raised to the 4th
3:18is the same as 2 x 2 x 2 x 2. Notice
3:23here that the 2 is appearing four times
3:25, which is exactly the number that is
3:27up here. Now, if the definition of
3:30exponentiation means multiplying the
3:33number by itself the number of times
3:35shown up here, see that j cubed
3:37corresponds to j x j x j. And how many
3:41times is j being multiplied here? 1, 2,
3:443 times. Therefore, by the definition
3:48of exponentiation, it necessarily needs
3:50to be three. So, we have already
3:52discovered one of the values of the
3:54letters in this item. Now we have a
3:57number that raised to the power of 3
3:59results in A, because these are all
4:01equalities. A is a single-digit number,
4:05because Julia uses these letters to
4:07represent the digits from zero to nine.
4:10So, a number raised to the power of 3
4:13corresponds to a single-digit number.
4:16In other words, if I take a
4:18single-digit number and multiply it by
4:20itself three times, it remains a
4:21single-digit number. Let's test which
4:24possibilities we would have. Clearly,
4:26since multiplying three times keeps the
4:28value small, a single digit only. This
4:31J must be a small number, but it cannot
4:34be zero or one, because zero and one
4:36are already the values of the letters B
4:38and F, as we saw in item A, because
4:40originally, if I put 0³, it would give
4:43a single-digit number, which would be
4:45zero, or 1 cubed, it would also give 1,
4:47which would work. So, the next digit we
4:50can test is two. Will it work with 2?
4:542³ corresponds to 2 x 2 x 2, which is
4:588. So we have that when I set j = 2,
5:01the result remains a single-digit
5:04number. If I took a number greater than
5:07two, it would already give a result
5:09with a number that has two digits. For
5:11example, 3³ is 27. In short, we have
5:14just discovered that the value of J
5:17must necessarily be two and,
5:19consequently, the value of A
5:21corresponds to 8, which is the value of
5:242³. And this concludes this item.
5:27Julia joins letters together to form
5:29numbers with two or more digits. For
5:33example, she correctly writes that JC +
5:35A = CF. Let's first interpret what that
5:38means. We have just realized that Julia
5:41can take these letters and place them
5:43side-by-side to form a number that has
5:45more than one digit. From the previous
5:48item, we just saw that J equals 2 and
5:51that C equals 3. In other words, this
5:54means that by placing J next to C,
5:56Julia can represent the number 23. And
6:01applying this same reasoning to the
6:03rest of this expression, see that Julia
6:06can encode the expression 23 + 8 = 31,
6:08by placing the letters exactly in the
6:11form JC + A = CF. Knowing this
6:13information, let's move to item C. To
6:16wrap up, Julia sets up the calculation
6:18below. Which digits do the letters D, I
6:22, E, H, and G represent? Now we have a
6:25huge calculation with a bunch of
6:27letters. And for us to be able to
6:29understand how to solve this question,
6:31we first need to get organized. Let's
6:33start by taking the letters whose
6:36values we already know from items A and
6:39B. So, we know that A equals 8, B
6:42equals 0, C equals 3, F equals 1, and J
6:46equals 2. In the set-up calculation,
6:50the letters that appear among these are
6:52J, C, and F. Let's replace each one
6:55with its respective value. So, in the
6:58calculation set up here, we already
6:59have some numbers discovered. Now that
7:03we have managed to place the numbers
7:05whose values we know, let's try to
7:06discover the other numbers that
7:08represent each of the letters based on
7:10this configuration here. Looking here
7:14at the top, we have that 220-something
7:16times 31 will give us the rest of this
7:18calculation. For me to perform this
7:21multiplication, first I will do here,
7:23look: 1 x d will give d, 1 x 2 will
7:26give 2. 1 x 2 will give 2. For now, I
7:28haven't managed to discover anything.
7:30Now when I move to the 3, I will do 3 x
7:33d and the result will be d. So this
7:37means that when I take 3 and multiply
7:39it by the single-digit number d, the
7:42result will be something that ends in d
7:44. What are the possibilities for this
7:47to happen? Well, I could put 3 x 0 to
7:49end in zero, but I already used zero,
7:51which is the value of b. The same thing
7:54with 1. 3 x 1 is 3, but I already used
7:57it for the value of f. Now, if I think
8:00about the other numbers, 3 x 2 is 6. So
8:02it doesn't continue ending in 2 here.
8:05We already used 0. And 4 too. 3 x 4 is
8:0812. It ends in two. We already used 2.
8:12And then when we reach 5, we will
8:15observe that 3 x 5 = 15, that is, the
8:19number ends in five. So I can replace
8:22the d. x = 15. And then one goes up
8:29here. And then I could keep the value
8:32of d. So, we just found out that D is
8:35worth 5. By substituting these numbers
8:37in their respective places, we have
8:39already discovered the first letter. D
8:41is worth 5. Now let's continue the
8:44calculation. 3 x 5 = 15. So one goes up
8:47here, and 5 stays. Now 2 x 3 is 6. 6 +
8:521 will give 7. So the number 7 must go
8:56exactly in the place of h. Thus, we
8:59just found out that h is 7. Let's
9:01substitute it into this written
9:03calculation and put the value of h here
9:05. Now that I already know that h is
9:08worth 7, continuing the calculation, 3
9:10x 2 is 6. So here i must be worth
9:13exactly 6 for this calculation to work.
9:16Let's substitute i = 6 here and
9:18complete it down here where i also
9:21appears. Only the value of e remains to
9:23be discovered. Look here, continuing
9:26the calculation, what I will do now is
9:28add these two values here. It will be 5
9:31plus nothing is 5. 2 + 5 = 7. 2 + 7 is
9:369. So that means the value of e for
9:39this written calculation to be correct
9:41must be exactly that. Now then, by
9:44placing the value of e, we have already
9:46discovered the total of the written
9:48calculation, which is 225 x 31,
9:50resulting in 6975. This allows us to
9:55find out that D is 5, I is 6, E is 9,
9:58and H is 7. However, I don't know what
10:00the value of G is. G does not appear in
10:02the written calculation anywhere. How
10:04am I going to find out the value of G?
10:07Remembering that Julia coded each of
10:09the digits from 0 to 9 with a letter.
10:12We have already used a total of nine
10:15letters here to represent nine digits.
10:18But since there are 10 digits from zero
10:20to nine, that means only one remains
10:22that hasn't been used. And looking here
10:25, we have already used 0, 1, 2, 3. The
10:28one that doesn't appear is four. That
10:31means the only value G can take is the
10:33digit that Julia didn't use at any
10:35point in this question, which is four.
10:38Therefore, G equals 4. And finally, we
10:41have finished this item. D is 5, I is 6
10:44, E is 9, H is 7, G is 4. From the
10:48previous items, we already knew that A
10:51was 8, B was 0, C was 3, F was 1, and J
10:54was 2. And we discovered all the digits
10:57and all the letters that these digits
10:59represent for Julia. And this completes
11:02this question. A game starts with a
11:08list of numbers from 1 to nine, as
11:10shown in the figure. In each step of
11:14the game, a smaller list is formed from
11:16the previous larger list as follows:
11:18any two numbers are chosen and removed
11:21from the list. Subsequently, after
11:25these numbers are removed, the
11:27difference between the larger and the
11:29smaller of them is written at the end
11:31of the new list. If the numbers removed
11:34are equal, zero is written at the end
11:37of the list. The game ends when only a
11:40single number remains. The figure shows
11:44an example of three steps of this game.
11:47Item A: After the first step, what is
11:49the largest number that can be written
11:51at the end of the list? We have already
11:55realized that starting from the initial
11:57list, to perform step one, we need to
11:59choose two numbers and subtract them.
12:03Since we want the largest possible
12:05number, it makes sense to choose the
12:07largest and the smallest number from my
12:09initial list to perform the subtraction
12:12. Looking at the list of numbers from
12:15one to nine, I can see that the largest
12:18of them is exactly the one at this end
12:20here, which is nine. And what is the
12:23smallest number in this list? It is the
12:25one exactly at the other end, which is
12:28one. After all, this list contains the
12:30numbers exactly from 1 to nine. Thus,
12:33to maximize the value of the number
12:35that I will place at the end of the
12:37list after performing the first step, I
12:39just need to subtract the smallest from
12:41the largest number in the initial list,
12:43in this case, 9-1. And the largest
12:46number that can be written is 9-1,
12:48which is 8. Therefore, this is the
12:52largest possible value to place at the
12:54end of this list. Item B. After the
12:58first step, can the list have exactly
13:00four odd numbers? Justify your answer.
13:04Let's observe how many odd numbers we
13:06have in our list after each step.
13:09Initially, before performing step one,
13:11we have five odd numbers in the list,
13:14right? 1, 3, 5, 7, and 9. Let's write
13:19this in the corner so we don't forget.
13:21Before step one, we have five odd
13:23numbers. Now, after performing step one
13:26, following the sequence described here
13:28in the notebook, let's see how many odd
13:30numbers are left. We have 1, 2, and 3
13:33left, right? 1, 5, and 9. So now we
13:36have exactly three odd numbers. After
13:40performing one more step, notice that 1
13:43, 5, and the other one remain, so there
13:45are three odd numbers. In the next
13:49stage, after the following step, 1, 5,
13:51and the other remain, so there are
13:53still three odd numbers. What can we
13:57observe in common after performing each
13:59of these steps? Look here, the amount
14:02of odd numbers I have after performing
14:05each step is always odd. Here it was
14:08initially five, then it went to three,
14:10and stayed at three. And will it always
14:13stay at three or will it decrease, but
14:15could it decrease to two? Let's see.
14:18Let's try to think about a certain step
14:20four. When I'm taking step four, those
14:23numbers can be even or odd. Look here,
14:27if I chose two odd numbers, then I'll
14:29remove two odds here, suppose 5 and 3,
14:31I removed those two odds, and I will
14:34replace them with the difference.
14:36However, the difference between two odd
14:39numbers is always even. So only one odd
14:42number will remain, which is one. On
14:44the other hand, if I had the difference
14:47between an even and an odd number, the
14:49difference will always be an odd number
14:51. Look up here. 9-0, an odd minus an
14:54even, will result in an odd. So, the
14:57quantity of odd numbers will be
14:59maintained. So, it either stays the
15:01same or it decreases by two. Since the
15:05initial quantity of numbers was odd,
15:07then this quantity will always be odd.
15:10And four is an even number. Therefore,
15:13it is impossible for us to have exactly
15:16four odd numbers, because the number of
15:18odd numbers in the list is always odd.
15:22In summary, the answer is no, because
15:24the quantity of odd numbers in the list
15:26after each step is always odd,
15:28precisely because of these little rules
15:30here. Even minus even is even, even
15:33minus odd is odd. Odd minus even is odd
15:36. Odd minus odd is always even. Item C.
15:39Explain why the number that remains at
15:42the end of the game is always odd.
15:44Let's adopt the same strategy we
15:46thought of in the previous item,
15:47thinking about the parity of the
15:49numbers. We already know that in each
15:52of the lists after the step, the
15:54quantity of odd numbers is always odd,
15:56precisely because of these little rules
15:59here, that when I subtract two even
16:01numbers, the result is even. Even minus
16:04odd gives odd, odd minus even also
16:06gives odd. Or if I subtract two odd
16:09numbers, the result will be even. So,
16:12after I perform any number of steps, I
16:14know that since there is a quantity of
16:17odd numbers left, I will never be able
16:20to end up with an even quantity of odd
16:23numbers. And at the end of the game,
16:26exactly one number will remain. Well,
16:29one is an odd number, so I would need
16:32to ensure that at this stage of the
16:35game I had an odd number of odd numbers
16:39. If I only have one number, this
16:42number will need to be odd, because I
16:44always need to have either five or
16:46three or one. I must always have an odd
16:49quantity of odd numbers. And if this
16:52number were not odd, it would be even.
16:55But if this remaining number is even,
16:57in this list I will have zero odd
16:58numbers, and I cannot have an even
17:00quantity of odd numbers. Thus, we have
17:03just discovered that since at each step
17:06there is always an odd quantity of odd
17:08numbers, it is impossible that at least
17:10one odd number does not remain after
17:12any step, because this quantity of odd
17:14numbers cannot be even. And if the
17:16number that remained were, for example,
17:18two, then the quantity of odd numbers
17:20that I have in this list, which only
17:22has the two, is zero. And this cannot
17:24happen. Therefore, the number in the
17:26last step must be odd. And this
17:28justification concludes this question.
17:34Each of the three squares below has
17:37sides measuring 6 cm and was divided
17:39into regions of equal area. Item A.
17:42What is the length X? Observe that the
17:46square was divided exactly into four
17:48figures, being two central squares here
17:51and two rectangular strips here in the
17:53corners. The problem statement told us
17:56that the side measurement of the square
17:59is 6 cm. See then that since we have a
18:03square, the side measurements of a
18:04square are always the same. Therefore,
18:07if this measures x, it means that this
18:09piece here also measures x. And by the
18:11geometry of the question, we know that
18:13this bottom part is a square equal to
18:15this one here. So, this piece here also
18:18measures X. Since this here is a
18:20rectangle, it means that the sum of
18:22these two segments corresponds to 6 cm.
18:26Thus, this means that 6 corresponds to
18:292x and so the x is equivalent to 6/2,
18:32which is 3 cm. And this is the value of
18:37length x, 3 cm. And this concludes this
18:40item here. Item B. What is length Y?
18:44Now the figure has been subdivided into
18:47three rectangles. We know that this
18:50square has a side measuring 6 cm. And
18:54that means the total area of the square
18:57is 6² = 36 cm². Since the regions
19:01into which the square was subdivided
19:04have the same area. And here we have
19:07three regions. Each of these regions
19:10measures 36/3, which corresponds to 12
19:15cm². Knowing this, observe that down
19:20here we have that this measure is 6 cm,
19:23but it also corresponds to these two
19:26halves, and each half is a side of the
19:29rectangle. So we know that since this
19:33piece here is equal to this one and the
19:35sum of these two pieces is six, this
19:37little piece measures 3 cm and this one
19:39also measures 3 cm. But we already know
19:44that this rectangle here has a total
19:46area of 12. After all, each of these
19:51parts has exactly 12 cm² of area.
19:54Therefore, to calculate the area of
19:57this rectangle, I do 3 x y, I do the
19:59length times the width, and that equals
20:0212 cm². Thus, we just discovered that
20:073y = 12. And solving this equation, we
20:11discover that y is 12/3, which equals 4
20:14cm. So, we just discovered that length
20:19Y is 4 cm, which concludes this item B.
20:22Item C. What is length Z? Now we have
20:27that our square has been subdivided
20:30into one rectangle here, two trapezoids
20:32, and a triangle. Let's gather once
20:36again the information we already know
20:38about this situation here. Since the
20:42side of our square measures exactly 6
20:45cm, its area is 6² = 36 cm². We also
20:51know that each of these parts here into
20:54which the square was subdivided has
20:56equal areas. Thus, since we have four
21:01regions, this means that each one
21:04measures exactly 36/4, which
21:06corresponds to 9 cm². Let's now take a
21:11look at the two trapezoids in this
21:14division of our square. Let's look at
21:17this trapezoid down here. How do we go
21:21about calculating the area of a
21:22trapezoid? If we have a measurement of
21:25a base, let's consider the trapezoid
21:27looking from below, that this measure
21:29of one base is the larger base. This
21:32small piece here is the smaller base.
21:35This part here is the height. To
21:38calculate the area of a trapezoid,
21:41let's say that for this green trapezoid
21:43here, the smaller base is b1 and the
21:45larger base is B1. We need to add the
21:50measurements of the two bases, that is,
21:54take big B1 + small b1, multiply the
21:57whole value by h, and divide the result
22:01by 2. This will be the formula for the
22:05area of this green trapezoid. We
22:09already know that the value of this
22:12area corresponds to, consequently, this
22:16means that big B1 + small b1 times this
22:20height h must be equal to 2 x 9, which
22:24is 18. But now we have three variables,
22:29right? Big B1, small b1, and h. And we
22:32haven't the slightest idea what the
22:34value of each one is. So let's look at
22:37the other trapezoid in this figure to
22:40see if we can get more information. Now
22:43that we have two pieces of information
22:46relating the bases and heights of these
22:49two trapezoids, let's try to find a way
22:52to relate their base measurements,
22:55precisely to try and find a way to
22:57determine the value of Z. What do we
23:00know? Notice that according to the
23:05figure, we know that the sum of the big
23:09base B1 + the big base B2 must be
23:12equivalent to 6. After all, this
23:16segment, this piece here, this region
23:18is precisely a rectangle. So, if this
23:21piece measures six, it means that this
23:23one measures six as well. And in the
23:26same way, we know that if I add small
23:30b1 with my value of Z, which I want to
23:34discover, with my small b2, this sum
23:37here must also result in 6. I had also
23:42already discovered that big B1 + small
23:45b1, if I move the h to divide,
23:48corresponds to 18/h, which is the same
23:51as big B2 + small b2. If we could find
23:55the value of h, we would know exactly
23:57what the sum of big B1 with small b1,
24:00and big B2 with small b2 is. To find
24:04this value, let's look at this
24:06rectangle here. This rectangle contains
24:11exactly two side measurements, one
24:13being six and the other measuring h.
24:18This, in other words, means that the
24:20area of this rectangle is 6 x h. But
24:24this rectangle corresponds exactly to
24:27the sum of the area of three regions of
24:30my larger pink square. Since each of
24:33these regions we had already seen is
24:36worth 9 cm² in area, it means the
24:39total area of this rectangle is 3 x 9,
24:4227. From that, we can discover that the
24:46value of h/6, which is 4.5. And if h is
24:544.5, it means that this sum of big B2
24:57with small b2 and big B1 with small b1
25:01corresponds to 18/4.5, which is the
25:04same as 36/9, which corresponds to 4.
25:09So we have now discovered the value of
25:12these two sums. Our goal with the
25:15question is to find the value of Z.
25:17Note that here I have the sum of the
25:20big B1 with big B2, small b1 with small
25:22b2, and with z. If I add these two
25:28equations here, I will get the sum of
25:31big B1 with big B2 plus small b1 plus
25:35small b2 plus the value of Z. All this
25:39results in 6 + 6, which is 12. But up
25:44here, I also already know how much big
25:47B1 + small b1 is worth and how much big
25:49B2 + small b2 is worth. If I substitute
25:52here, only the value of Z will be
25:55missing. So look, big B1 + small b1 is
25:58worth 4. And big B2 + small b2 is also
26:02worth 4. This means, in other words,
26:06that the sum of big B1 + big B2 + small
26:09b1 + small b2 is 2 x 4, which is 8.
26:13This means that 8 + z corresponds to 12
26:17. And in this way, we have just
26:21concluded that Z is 12-8, which is 4 cm
26:24. And this is the value of the length Z
26:28that we wanted, which finishes this
26:29question here. Gabriela covers
26:36triangular boards and rectangular
26:39boards with L-shaped pieces, formed by
26:42three small squares equal to the
26:44squares of the board, without
26:46overlapping. She has enough pieces to
26:51cover boards of any size. So, we have
26:55here the L-shaped pieces and some of
26:58the boards that Gabriela assembled, a
27:00square board, a rectangular one, and a
27:03triangular board. And it asks to show
27:07how Gabriela can cover a triangular
27:08board with a base of eight. First, note
27:12that it is called base eight because
27:15here at the bottom I form the board
27:17with exactly eight small squares. The
27:21problem statement already shows us how
27:24to fill the triangular board, but with
27:27a base of six. So, I already know how
27:30to do this part here. I can then think
27:33about how to fill and complete the one
27:35with a base of eight. So, if I take the
27:39one with a base of six, placing it here
27:42on top, note that exactly these two
27:45rows down here will be missing. And we
27:48also already know how to assemble a
27:51rectangular row. The problem statement
27:53also shows us this. And if we join it
27:56with an L-shaped piece here in the
27:58corner, we will be able to make these
28:00two missing rows. So, with the
28:04triangular board of side six, we can
28:06assemble this upper part in red. With
28:10the rectangle that Gabriela made, we
28:12can assemble this part in blue. And to
28:16complete the triangular board of base
28:18eight, just add an L-shaped piece here
28:20in green. And this is one way to fill
28:24Gabriela's board as we wanted. Item B.
28:28Explain why Gabriela will never be able
28:30to cover a triangular board of base
28:33seven. We have already seen how to fill
28:36a triangular board of base six. The
28:39figure is already in the problem
28:40statement for us. If I want a figure of
28:43base seven, notice that I would need to
28:46add one more row down here. However,
28:50our L-shaped piece, regardless of the
28:53position I place it in, will always
28:55occupy two rows or two columns of my
28:58board figure, because it has those two
29:01little ends, one pointing here and the
29:04other pointing there. So, regardless of
29:08how I set up the configuration, you
29:11could say to me, right? But I could not
29:13use the initial configuration that was
29:15given in the statement. I could, for
29:17example, take an L-shape and do it from
29:19the bottom up here. So, if I did that,
29:22notice that at any moment I would
29:24always have something missing; since I
29:27have an odd number of squares here, I
29:30would always only be able to complete
29:32an even number or multiples of three.
29:36Since seven is neither even nor a
29:38multiple of three, I cannot complete it
29:40if I think from top to bottom. That is
29:44why in this case I cannot complete the
29:47triangular board if it has seven rows.
29:52But besides this geometric
29:53justification, notice also that there
29:55is another justification that is even
29:57easier, an arithmetic one. A triangular
30:01board that has base seven will have a
30:03total, if we add one square in the
30:06first row, two in the second, three in
30:09the third, and so on, totaling 28
30:11squares. And the L-shaped pieces that
30:15Gabriela has, all of them have exactly
30:18three squares. Oops, 28 is not
30:21divisible by three. So it is impossible
30:25for us to cover the board using only
30:27L-shaped pieces, where each one of them
30:29occupies exactly three squares. If I
30:32completed as much as possible,
30:35eventually I would complete 27 squares
30:37and one would be missing, precisely
30:40because 28 is not divisible by 3. So
30:43this is another justification that is
30:45based on the number of squares that the
30:47triangular board of base seven has. And
30:50on the number of squares that the
30:52L-shaped figure Gabriela uses also has.
30:56Both justifications are valid for
30:58answering this item. Item C. Explain
31:01why Gabriela can always cover a
31:03triangular board with a base of 6k for
31:06any k equal to 1, 2, or 3. Let's try to
31:10interpret this formula first. When I
31:12talk about a board with the formula 6
31:15times k, I mean that this value of k
31:18can take on any of those numbers. So,
31:20if k equals 1, I'm thinking of a
31:22triangular board with a base of 6. If k
31:25equals 2, I'm thinking of a board with
31:28a base of 6 times 2, which is 12. If
31:31the base is 3, I think of a base of 6
31:33times 3, 18. And so on. We've already
31:38seen in the problem statement how we
31:39are given how to assemble the
31:41triangular board with a base of six.
31:45Now, can we assemble a board with a
31:46base of 12 starting from the base-six
31:49board? In the statement, besides the
31:53base-six board, we are also shown how
31:56to cover a square board with a base of
31:58six. So, if I take this square board
32:02with a base of six and join it with two
32:04triangles, two triangular boards with a
32:07base of six, couldn't I form a
32:09triangular board with a base of 12?
32:12Look at what will happen. We will then
32:15have two triangular boards here with a
32:17base of six, which I've placed here in
32:20red, and one square board with a base
32:22of six here in green. And together,
32:26they will form a triangular board whose
32:29base measures six here and six here as
32:32well, resulting in 12. Thus, from a
32:36base-six board, I can assemble one with
32:39a base of 12 just by joining two
32:41triangular boards with a base of six
32:44and one square board with a base of six
32:47. I already know how to assemble those
32:50two. The statement itself shows me how
32:53to do this using the L-shaped pieces.
32:56But what if I wanted to make the
32:57triangular board with a base of 18 now?
33:01I just need to follow the same tune. I
33:03take two triangular boards with a base
33:06of 12, which I now know how to make.
33:09And for what is missing, I fill it in
33:12with square boards with a base of six.
33:15In the case of 12, I will need to use
33:17exactly four of them. And following
33:20this same reasoning, I have the
33:22following method to fill a triangular
33:24board with a base of 6k. When I am at
33:27step k, I just need to join two
33:29triangular boards with a base of 6 (k-1
33:32) to make the corners. And in the
33:36middle, I join (k-1) ² square boards
33:38with a base of six. The circles in the
33:45diagram below must be filled with the
33:48numbers 1, 2, or 3, one number in each
33:51circle, so that circles connected by a
33:53segment have different numbers. The
33:57figure to the side shows one possible
33:59way to fill the diagram. We have here
34:02the diagram with the five circles
34:03filled with numbers according to the
34:05problem's constraints. Let's move on to
34:07the items. Item A: Complete the diagram
34:11below with a possible correct matching.
34:14Let's first recall what the rules are
34:16for filling in this diagram. The
34:20problem tells us that whenever there is
34:22a line connecting two circles, these
34:25two circles must have different numbers
34:27. Furthermore, we know that each circle
34:32must contain exactly one number between
34:351, 2, and 3, only these. Thus, if, for
34:39example, in this connection that I
34:41filled in red, if I have the number one
34:44at the top, it means I cannot have one
34:47at the bottom; I can only have two or
34:50three, because circles connected by a
34:52segment cannot have the same numbers.
34:56But besides that, this ball here in the
34:59center is also connected to the top
35:01ball here, which has the number two
35:04fixed. This means that this ball cannot
35:07receive the number one nor the number
35:09two. Now, since we must fill it with
35:13the numbers one, two, or three, only
35:14the number three is left for me to
35:16place here. And note that the same
35:20reasoning applies to all the balls here
35:22at the bottom. Look here, for example,
35:26at this ball; it is also connected to
35:27the ball with number one and the ball
35:29with number two. So here it can only be
35:33three, and repeating the same tune on
35:36this end, it connects to ball two and
35:38connects to ball one. So here we can
35:42only place the three. So we almost have
35:45our entire diagram filled out. This
35:47ball here is missing. So this ball
35:50cannot have the same number as any
35:52connection. Since the connection it is
35:55making is with these three bottom balls
35:58, which we know necessarily need to
36:00have the digit three, we can choose
36:02either one or two to place here. This
36:07means that to complete the diagram, we
36:09have these two possibilities here. In
36:13both, the bottom part must be three. In
36:16this circle up here, we can put one or
36:18two. And this finishes this item here.
36:22Item B. In how many ways is it possible
36:24to correctly fill out the diagram below
36:27? Let's adopt the same strategy we
36:29thought of in item A. Initially, we
36:31have three numbers to place. One, one,
36:34two, or three. Imagine, for example,
36:36that I am going to place the one here.
36:39According to the rule, we know that the
36:41number here and the number here cannot
36:43be one either, because there is a
36:45connection between the two circles. So,
36:48neither here nor here can it be one.
36:50Therefore, I can put two or three. In
36:53what order can I place them? It will
36:55depend on the number I place here. If
36:58this number is also one, let's start by
37:00thinking about that. If the two numbers
37:03up here are equal, them being equal
37:06guarantees that I will have the freedom
37:08to choose the others down here. Because
37:11look, if, for example, I chose one here
37:14and one here, I can put either two or
37:16three here. Or I can put three or two
37:20here. I can also put just the two and
37:22leave the three out. Or also put just
37:25the three and leave the two out. I will
37:28have two possibilities to choose a
37:30number here, which can be two or three,
37:33and also two possibilities to choose
37:35the two or three here. total. So I will
37:38have 2 x 2, four possibilities if I
37:42choose the same number to put here. In
37:45this case, I chose one, but I could
37:48choose two to put here. If I put two
37:50here, then it will be the same story
37:52here. The only change is that I will be
37:54able to put one and three, or one and
37:56three here. This means that if the two
38:00numbers up here are equal, for each
38:02choice of two equal numbers up here, I
38:05have four possibilities to choose
38:07numbers down here. So, what numbers can
38:11I choose up here? one, two, and three.
38:14So I have three options here below.
38:16Then I can put the remaining number
38:18here. For example, if I put one here,
38:21two and three, here two and three, we
38:22have a total of four possibilities here
38:24. Since these choices are independent,
38:28we have a total of 3 x 4, which equals
38:3112 possibilities, if the two numbers up
38:34here are the same. However, what
38:37happens if the numbers here are not the
38:40same? Let's calculate the possibilities
38:42with different numbers. If we have
38:45different numbers up here, for example,
38:47let's start by thinking about one and
38:49two. Notice that since each circle down
38:52here is connected to the two above, see
38:55, this one is connected to this one and
38:57this one. And this one is also
38:59connected to this one and this one. The
39:00numbers down here cannot be one or two.
39:03So what will be left? the three for it.
39:05And here, only the three will be left
39:07as well. This means that given a choice
39:11of two distinct numbers up here, the
39:13choice below is fixed. So, I just need
39:17to count how many ways I can make the
39:19choice up here. So, for each choice of
39:22two numbers to put up here, for example
39:24, one and two, I have two possibilities
39:26. I can put one here and two here, or I
39:29can put two here and one here. And so,
39:31what choices do I have? I can put one
39:34and two, I can put two and three, or I
39:36can put three and one. I then have
39:39three choice options here. For each of
39:42these three options, I can place them
39:44in two different orders. In total, then
39:47, I have 3 x 2, which is six
39:49possibilities, if the top numbers are
39:52different. Therefore, we conclude that
39:55in total there are 12 possibilities to
39:57fill this diagram with identical
39:59numbers on top. six possibilities with
40:03different numbers on top, which results
40:05in 12 + 6, 18 possibilities. And these
40:09possibilities can be listed here. Note
40:12that in these first 12 options here, I
40:15have the top numbers as identical, and
40:17in these other six options, the top
40:20numbers are different. And these are
40:22the 18 possibilities to solve and fill
40:25in the diagram presented in the item.
40:28Item C. In how many ways is it possible
40:31to correctly fill in the diagram below?
40:34Let's start with the simplest case,
40:36pretending that all the numbers in the
40:38sequence above are the same. For
40:41example, if we place only ones in these
40:44three circles up here, look below: I
40:46can no longer place a one, but I am
40:49free to place either two or three in
40:51any of these four circles. A two or
40:54three can go here, a two or three here,
40:57and so on. This means that I have two
41:00possibilities for this circle, two for
41:02this one, two for this one, and two for
41:04this one in total. This means I will
41:08have 2 x 2 x 2 x 2, which is 2 to the
41:11power of 4, that is, 16 possibilities
41:15for each choice of identical numbers I
41:18make up here. Wait, I can choose here 1
41:231 1, 2 2 2, or 3 3 3; here I have three
41:26possibilities, and each of these three
41:29possibilities generates 16 fillings.
41:33Therefore, the total number of
41:35possibilities in this case is 3 x 16,
41:37which is 48. Indeed, we can set up a
41:40small table where we place the sequence
41:43above with identical digits, right?
41:46Here we have 1 1 and the respective 16
41:48sequences at the bottom for each of the
41:51selected top sequences. In this table,
41:55it is much easier to see that the total
41:57number of possibilities is three,
42:00because I have three top sequences
42:02times 16 possibilities for bottom
42:04sequences, which equals 48. Now that we
42:08know what happens when the sequences
42:10have the same numbers. Let's think
42:14about what occurs when I have at least
42:16two different numbers at the top. In
42:19this situation, imagine I place a one
42:21here and a different number here in the
42:23middle, a two, for example. Since each
42:26of the four bottom circles is connected
42:28to these two circles, I cannot place a
42:31one or a two here. Thus, only the three
42:34remains to be placed. This means that
42:37once I have chosen two numbers up here
42:39in this part, the bottom number will be
42:42fully defined as the only number that
42:44has not yet appeared. So, given a
42:48choice of two numbers to place here, we
42:50will have only one possibility for the
42:53bottom sequence, which is the only
42:55number that did not appear at the top.
42:58We can place ones and twos in these
43:00three circles here. Thus, we have a
43:03total of 2 x 2 x 2, which corresponds
43:07to 2³, which is eight possibilities.
43:10However, see here in this count, I am
43:13considering the positions where I have
43:161 1 1 and 2 2 2. These are situations I
43:19have already calculated in the previous
43:22case. So I need to subtract these two
43:24cases where I have only one number,
43:27which would be 1 1 1 or 2 2 2. In this
43:30way, this means there are a total of
43:33six possibilities for matching the
43:35diagram once I have chosen two numbers
43:38from one to three to fill the top. And
43:41in how many ways can I fill this top
43:43part? Well, if I have the numbers one,
43:47two, and three, I can choose either
43:49numbers one and two, or choose two and
43:52three, or even choose one and three. In
43:56total, we have three possibilities.
43:58Each of these three possibilities for
44:00choosing the numbers here in the top
44:02sequence will generate six fillings.
44:05Thus, the total number of possibilities
44:07is 6 x 3 = 18. And in the same way, as
44:11we did before, we can set up a table.
44:14Notice now that it is the bottom
44:17sequence that will have all the circles
44:20with the same numbers, either 1 1 1 or
44:242 2 2 or 3, and in total, therefore, 3
44:28x 6 possibilities, 18 as we had
44:31observed. Now, it remains to check what
44:34happens if I were to place three
44:36different numbers at the top. For
44:38example, 1, 2, and 3. In this situation
44:41, it would be impossible to fill in the
44:43rest of the diagram, because down here
44:45I wouldn't be able to put 1, 2, or 3.
44:48And the problem statement forces us to
44:50use one of these three digits to fill
44:52in the circles. Thus, there are no more
44:55possibilities left for us to check.
44:57Therefore, this means we have 48
45:00possibilities if the top numbers are
45:03equal, plus 18 possibilities if we have
45:06two equal numbers on top and one
45:09different. Or, in other words, if the
45:12numbers on the bottom line are equal.
45:15And in total, this results in 48 + 18,
45:19which is 66 possibilities. A set of
45:26non-zero digits is called a "quadruplex
45:29" when the quadruple of one of its
45:31elements is equal to the sum of all the
45:33others. For example, 2 3 5, 2 4 6 8,
45:38and 1 2 3 4 5 6 7 8 9 are quadruplexes.
45:43Item A. Find a quadruplex set with five
45:46elements. We first need to observe that
45:50a quadruplex set is composed only of
45:53single-digit numbers. Thus, we know it
45:57is composed of digits from 1 or 2 or 3
46:01up to 9. So the largest quadruplex we
46:04can have is the very one given in the
46:07problem, which has exactly nine
46:10elements. Let's now understand what
46:12this quadruplex property actually is.
46:14For example, 2 3 5 is a quadruplex
46:17because if I take 2 and multiply it by
46:204, it equals the sum of 3 and 5. Here
46:23it is the same thing. If I choose 4 to
46:26multiply by 4, it equals the sum of 2,
46:296, and 8. So, to think of a quadruplex
46:33set with five elements, I could take,
46:36for example, the first five natural
46:38numbers and test if they form a
46:40quadruplex set. This is a quadruplex
46:46set, because I can take, for example,
46:49the element 3 and calculate 4 x 3,
46:53which is 12, and 12 corresponds exactly
46:57to the sum of 1 + 2 + 4 + 5, which are
47:00the other four elements of this set. So
47:05this one, 1 2 3 4 5, is an example of a
47:08quadruplex set with five elements, but
47:11you could show many other examples. In
47:15total, we have 15 quadruplex sets with
47:17five elements, which are these ones
47:20shown here. The simplest one of them,
47:22obviously, is 1 2 3 4 5. Or we could
47:25also think in reverse and put 5 6 7 8 9
47:28. So, just choose your favorite
47:31quadruple-set from these 15. And it
47:34serves as the solution for this item.
47:38Item B. Explain why the sum of the
47:40elements of a quadruple-set is always a
47:42multiple of five. Let's take that
47:45example from our previous set. I chose
47:47a certain element. In this case, I
47:50picked three. Let's assume I don't know
47:53that it's three that I picked, and
47:55let's say this element I chose is X.
47:57What does the property of a set being
47:59quadruple tell us? If I take this
48:03number here, which is my x, and
48:05multiply this x by 4, the result
48:08corresponds to the sum of all the rest
48:11of the elements. So, if I add this one
48:15to this, this, and this, the result is
48:194x. But, what is the sum of these
48:22elements here? The sum of all elements
48:25would be equivalent to the total sum of
48:28the set, which I will call capital S,
48:31minus X, which was the one I chose to
48:33multiply by 4. Notice that in the
48:37example, the sum of the five numbers
48:38was 15. I chose x = 3. How much was
48:42left? 15-3 = 12. So, the sum would be
48:4615 and my x in this example would be 3.
48:49And 15-3, which is 12, is exactly 4 x 3
48:53. In other words, 4x is equivalent to
48:56the total sum of the numbers minus x.
48:59If I move x to the other side here,
49:02since it's subtracting, I move it
49:05adding, I will obtain that the sum of
49:08all numbers corresponds to 4x + x,
49:11which corresponds to 5x. And 5x is
49:16exactly a multiple of five. Therefore,
49:19we have just concluded that the sum of
49:22all numbers in the quadruple-set is
49:25five times something, that is, it is
49:28always a multiple of five. But this is
49:31not just any multiple of five; it is
49:33the multiple of five obtained with the
49:35number we chose to get the
49:37quadruple-set. In this case here, I
49:40chose three, which would be X. Item C.
49:43Find a quadruple-set with eight
49:45elements. As a quadruple-set, the
49:48maximum number of elements it can have
49:51is nine, which would be the
49:52quadruple-set from the prompt, which is
49:55precisely this one appearing here, to
49:57obtain a quadruple-set with eight
49:59elements, we just need to remove one of
50:02these digits here. However, we saw in
50:05item B that the sum of the elements of
50:08a quadruple-set is always a multiple of
50:11five. In this case here, by adding all
50:14the numbers from 1 to, I will obtain
50:16the sum 45. So I need to remove one
50:20digit here, and the resulting sum must
50:22continue to be a multiple of five,
50:25otherwise the set obtained will not be
50:27quadruplex. And among the digits from 1
50:31to 9, the only one that is a multiple
50:33of five is five itself. This means that
50:36starting from the quadruplex set with
50:39nine elements, to obtain one with eight
50:41elements, the only one I could
50:43eliminate would be five. Therefore, a
50:47quadruplex set with eight elements
50:49corresponds to 1 2 3 4 6 7 8 9, that is
50:52, just by removing five. And with this
50:56reasoning, we can actually see that
50:59this is the only quadruplex set with
51:01eight elements, precisely because the
51:03sum of all elements must be a multiple
51:06of five. Item D. Explain why there are
51:10only six quadruplex sets with seven
51:13elements. Imagine that from my set of
51:17numbers from one to nine, I will choose
51:19two of them to remove. So, suppose I
51:23choose A and B. The sum of all numbers
51:27in this quadruplex set will correspond
51:31to 45-a-b. But because this set is
51:36quadruplex, this means that this here,
51:39which is the sum of the elements of my
51:42set, actually corresponds to five times
51:45x, which is an element that is part of
51:49my quadruplex. In other words, if I
51:53move this A and B by adding them to one
51:56side, and subtract this five, the sum
51:58of the digits that I have to remove
52:01from the set of digits from 1 to 9, for
52:04what remains to be a quadruplex set,
52:06corresponds exactly to 45-5 x. Then,
52:12since 45 is a multiple of 5, I can
52:15factor out the 5 here, and write this
52:17as 5 times 9-x. Remembering here that x
52:22is an element of this quadruplex set
52:25that was not removed. So this X is at
52:29most, and since A and B are two
52:31distinct single-digit numbers, we know
52:34the maximum value for the sum of a + b.
52:41Which is obtained when I choose a = 8
52:43and b = 9 or vice versa. Knowing then
52:47that this here is at most 17 and also
52:50at least three, right? Because I can
52:53choose A = 1 and b = 2, this red
52:55calculation here on the right side must
52:58be at most 17. But 5 times 9-x will be
53:04a result that is greater than 17 if x
53:07is greater than or equal to 5. So this
53:12means that the digit x needs to be at
53:15least six. Therefore, we have four
53:17options. Either x is 6, or x is 7, or
53:21it is 8, or it is 9. But we also can't
53:25have that, because otherwise we'd have
53:279-9 here, which is 0. And since these
53:30digits are non-zero, the sum couldn't
53:32possibly be zero. So, we are actually
53:34left with only three cases to analyze.
53:38If x = 6, my sum a + b corresponds to
53:45exactly 5 x 9-6, which is 5 x 3, that
53:51is, 15. So, how can we add single-digit
53:56numbers to get a result of 15? If A is
54:016, for example, our digit B needs to be
54:049. So, the only option we have in this
54:09case would be 6 + 9. However, in a
54:12quadruplex set, we don't have repeated
54:15digits. So, since x here is already 6,
54:18this value here couldn't happen because
54:21a is already 6. Therefore, this
54:23situation here cannot occur. What is
54:26another way to get a sum of 15 with two
54:29single-digit numbers? We can have 7 + 8
54:32. And this one is allowed because 6, 7,
54:36and 8 are distinct digits. Now, if x is
54:42equal to 7, then a + b actually
54:46corresponds to 5 x 9-7, which is 5 x 2,
54:51which corresponds to 10. To get two
54:58single-digit numbers whose sum is 10,
55:01we can use 1 and 9, 2 and 8, 3 and 7,
55:04or 4 and 6. Remembering that A and B
55:08are distinct. And since X = 7 must be
55:13in my quadruplex, I couldn't have it
55:16and remove it at the same time.
55:19Therefore, this option is not valid. We
55:22are then left with the removal options
55:24of 1 and 9, 2 and 8, and 4 and 6. And
55:28finally, if x is equal to 8, following
55:32the same logic, we will have that a + b
55:36corresponds to 5 x 9-8. 9-8 is 1.
55:42Therefore, this is 5 x 1, which is 5.
55:47And in what way can we pick two
55:49distinct single-digit numbers whose sum
55:51is five? We can take 2 and 3 or we can
55:56take 1 and 4. Both options are valid
56:00because they don't coincide with the
56:02digit eight, which will already be in
56:04my quadruplex; moral of the story.
56:07Summarizing all this math we did, we
56:10discovered that we have one option if x
56:14is 6, three options if x is 7, and two
56:17more options if x is 8. So, in total,
56:22we have 1 + 3 + 2, which is six
56:25quadruplex sets with seven elements
56:28that we can form. And given the
56:32restrictions we studied, there is no
56:35way for a quadruplex set to exist that
56:37isn't formed from the removal of these
56:40sums here. So, we have these six
56:43possible sums here of numbers that we
56:45are going to remove, and each one of
56:48them will generate one of the six
56:50quadruplexes. Namely, we have these
56:53lists here with these six quadruplexes,
56:55which are the quadruplex sets with
56:57seven elements. And this concludes this
57:01question here. As a curiosity, we can
57:04see all the quadruplexes, from the
57:06smallest one, which has two elements,
57:07to the largest, which is in the problem
57:10statement, which has the nine digits.
57:13In total, we have 57 quadruplex sets
57:16that are shown here. M.

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