Category: Mathematics Competition

Take a look at the skyline: it features two separate sequences of towers. In each sequence, the towers progressively increase in height from left to right. An interesting detail is that the final two towers in both sequences make up the four tallest towers overall. (Although this happens here, it wasn’t a requirement—it could have been that all four tallest towers belonged to just one of the sequences.)

Given there are 14 towers altogether, if we set aside the four tallest ones, the next tallest tower would be ranked 10th in height. That tower is the one shown in Option A.

At the start of the week, Alba borrowed a book first, followed by Felix. Based on the original arrangement of the books, this means Alba took Charlotte’s Web, and Felix borrowed Curious George. Later, Cato borrowed a book right after Felix returned one. Since books are always borrowed from the top of the stack and returned to the top as well, Cato ended up borrowing the exact book Felix had just returned. Felix had only borrowed one book during the week—Curious George—so that must be the book he returned. Therefore, Cato borrowed Curious George. The true answer is option B.

Chest BEB stands out because it’s the only one where the first and third letters are the same. This chest corresponds to key 636, which is also the only key where the first and third digits are identical. From this, we can conclude that the letter B represents the number 6, and E represents 3. So far, we have:

• A = ?
• B = 6
• E = 3
• R = ?

Next, chest AER is unique because it’s the only one that ends with a letter other than B. It must correspond to key 934, which is the only key that doesn’t end in 6. Therefore, A corresponds to 9 and R to 4. Now the letter-to-digit matches are complete:

• A = 9
• B = 6
• E = 3
• R = 4

With these pairings, we can easily connect the remaining chests to their keys. For example, chest RAB pairs with key 496, which fills in the missing link. The true answer is option A.

Dani can start by organizing the bottles in order from the smallest capacity to the largest. At any moment, if there are two empty bottles with different capacities, it’s more efficient not to fill the larger one first. Instead, Dani should begin filling the bottles starting from the smallest and work upward in size. By doing this, the smallest seven bottles can be completely filled using 3 + 4 + 5 + 6 + 7 + 8 + 9 = 42 liters of water. After filling those bottles, there will be 8 liters remaining in the tank. However, since all the remaining bottles each require more than 8 liters to be filled, Dani won’t be able to fill any additional bottles completely. The true answer is option B.

First, we use the provided associations to figure out the special number for the red text.
Next, we calculate the special numbers for each of the texts listed in the answer choices and identify which one equals 6944.
Among all the options, the only text with the special number 6944 is the one shown in Option C.

Parts marked with “+” are available to everyone. To illustrate this, imagine duplicating each part labeled with “+” and placing a copy in every division and subgroup. For example, the Fox Subgroup has part 4, which is labeled with “+”. After cloning this part, the BIA would look like this:

Parts marked with “#” are accessible exclusively to the receiving division and its subgroups. We can replicate each part labeled with “#” and place a copy within the subgroups of the respective receiving division. For instance, the Wolf Division has part 2 labeled with “#”, so the Fox and Coyote Subgroups each get a copy of part 2. Similarly, the Squirrel Division holds parts 1 and 3 labeled with “#”, which means the Chipmunk Subgroup receives copies of parts 1 and 3.

Parts labelled with “=” are only accessible to the receiving division or subgroup so we don’t make clones of parts labelled with “=”. Looking at the BIA after all the clones have been made, we see that only the Chipmunk Subgroup has access to all four parts of the message. You should choose option D.

We label the peaks as shown.
Observe that the highest peak is $P_3$.

• If Binsa begins at $P_1$, she won’t climb any higher because both the peaks to her left and right are lower. Starting from $P_1$, she won’t be able to reach the highest peak. Likewise, if she starts on any peak to the left of $P_1$, she also won’t reach the highest peak, as she can’t move past $P_1$.
• If Binsa starts at $P_2$, she finds both neighboring peaks are taller, so she’ll climb to the higher one, $P_3$, and reach the highest peak.
• Starting directly on $P_3$ means she’s already at the highest peak.
• If she begins at $P_4$, she’ll climb up to $P_3$ and reach the highest peak as well.
• However, if Binsa starts at $P_5$, both adjacent peaks are higher, so she’ll climb to $P_6$. From $P_6$, she won’t climb any further because both its neighboring peaks are lower. This means she can’t reach the highest peak from there.
• Similarly, starting at any peak to the right of $P_6$ won’t lead her to the highest peak, since she can’t get past $P_6$.

Therefore, there are exactly three peaks from which Binsa can reach the highest peak: $P_2$, $P_3$, and $P_4$. You should choose option A.

In the provided image recording, the number 0 in the rightmost pixel of the middle row shows that no other images share this particular shade at that spot. Observing the images of the letters I, T, C, and L, they all display a white pixel in this location. However, the image representing the letter O features a black pixel instead. This distinct shade at the specified pixel, unique to the letter O, indicates that the image scanned by the machine must correspond to the letter O. Below are all the images along with their machine recordings.
Thus, you should choose option B. 

We can analyze each button and appliance separately. The key idea is this: if a button is pressed an odd number of times, the appliances it controls will end up in the “on” state. If a button is pressed an even number of times, those appliances will be left “off.” The same principle applies to each appliance individually—if it’s connected to an odd number of pressed buttons, it will be on; if connected to an even number, it will be off.

• Option A (T, R, Q, P): The vacuum cleaner is linked to three of these pressed buttons—T, R, and P. Since it’s connected an odd number of times, it will end up on, which doesn’t match the desired result.
• Option B (R, Q, P, S): The coffee machine is connected to just two of these buttons—P and Q. An even number of presses means it stays off, which isn’t correct.
• Option C (S, P, T, R): The washing machine is only linked to one of these buttons—S. A single (odd) press means it would be on, which is incorrect.

• Option D (Q, S, R, T): Here’s how the appliances are affected:

  • Laptop: connected to two buttons, Q and R → off (even).
  • Washing machine: connected to Q and S → off (even).
  • Television: connected to R, S, and T → on (odd).
  • Coffee machine: connected to Q → on (odd).
  • Vacuum cleaner: connected to R and T → off (even).

This option results in the correct on/off states for all appliances, making it the correct choice.

In Option C, observe that the final puzzle piece needs to have a flat right edge and be blue. This means the piece must be:
Additionally, the second-to-last piece should have a rounded tab on its right edge and also be blue. However, there are no blue pieces with a rounded tab on their right side—only square or triangular tabs are available in blue. Because of this, it’s impossible to complete the sequence described in Option C.

On the other hand, the sequences in Options A, B, and D can be successfully assembled as follows:

• Option A: YELLOW $\rightarrow$ BLUE $\rightarrow$ BLUE $\rightarrow$ RED $\rightarrow$ BLUE
  
  
• Option B: BLUE $\rightarrow$ YELLOW $\rightarrow$ RED $\rightarrow$ YELLOW $\rightarrow$ RED
  
  
• Option D: BLUE $\rightarrow$ RED $\rightarrow$ YELLOW $\rightarrow$ BLUE $\rightarrow$ RED
  

The image below illustrates how to assemble the craft at a total cost of 1 + 4 + 2 + 1 + 1 + 2 + 1 + 1 = 13 dollars. Since this is the lowest cost among the options provided, we can conclude that Option A is the correct answer.

To be thorough, let’s explain why it’s impossible to make the craft for less than 13 dollars. We can break the craft down into two main parts: the “head,” which can only be made from the three curved pieces, and the “body,” which includes the torso and legs and can be constructed from the remaining five shapes.

There are only two ways to create the head: either by using a single piece designed specifically for the head, which costs 6 dollars, or by combining the other two curved pieces, which cost 1 dollar and 4 dollars, totaling 5 dollars. So, the cheapest way to make the head is for 5 dollars. Now, let’s focus on the torso and legs. We need to show that the lowest possible cost to build these is at least 8 dollars (and not less than 7 dollars).

At the bottom left and bottom right of the torso, you must use either a trapezoid costing 7 dollars, a parallelogram costing 3 dollars, or a triangle costing 2 dollars. Let’s look at all the possibilities:

• If you use the trapezoid, you’ll still need additional pieces to complete the torso and legs, which would push the total cost over 7 dollars.
• If you use the parallelogram, you’ll also need to add two triangles, as shown in the example. However, even with these pieces, the legs won’t be complete, and the total cost will exceed 7 dollars.
• The last option is to place a triangle at each bottom corner of the torso, costing 2 + 2 = 4 dollars. You’ll then need to cover two remaining square sections—one at the top of the torso and the other forming the legs. The cheapest way to do this is by using four rectangles, which would cost an additional 4 dollars. This brings the minimum total cost for the torso and legs to 4 + 4 = 8 dollars.

Therefore, the least amount you can spend to build the entire craft is 5 dollars for the head plus 8 dollars for the torso and legs, which adds up to 13 dollars. You should choose option A.

First, Alpha delivered 20 tonnes of vegetables, splitting them evenly between Gamma and Beta—each receiving 10 tonnes. From there, 10 tonnes can be sent to Kappa via Delta. Now, focusing on Beta’s situation: we can’t send 2 tonnes to Gamma because the route from Gamma to Delta has a capacity limit of 10 tonnes, which has already been used. So, the only option is to send 3 tonnes from Beta to Delta and another 5 tonnes directly to Eta. Then, Delta sends its 3 tonnes, and Beta sends its 5 tonnes, both to Eta. This results in Eta having a total of 8 tonnes. Finally, all 8 tonnes from Eta are sent to Kappa. Adding that to the 10 tonnes already sent through Delta, Kappa ends up receiving a total of 18 tonnes of vegetables—the maximum amount that can be shipped.  You should choose option B.

Options A, B, and C can each be achieved in exactly three mutation steps:

• Option A: substitution (change T to C → GCATCG), substitution (change T to A → GCAACG), and another substitution (change C to T → GCAATG).
• Option B: substitution (change G to A → ATATCG), duplication (duplicate T → ATTATCG), and duplication (duplicate C → ATTATCCG).
• Option C: substitution (change T to A → GAATCG), substitution (change C to G → GAATGG), and substitution (change G to C → GAATGC).

Each of these sequences can be obtained with 3 gene mutation steps. However, Option D requires 4 steps to be formed: duplication (duplicate G → GGTATCG), duplication (duplicate A → GGTAATCG), substitution (change T to A → GGTAAACG), and deletion (remove G → GGTAAAC).

To understand why Option D can’t be reached in just 3 steps, consider the following reasoning:
Option D (GGTAAAC) has one more base than the original sequence GTATCG. This means there must be more duplications than deletions involved.

If we try to limit the process to 3 steps, there are only two possible scenarios:

• Two duplications and one deletion: To create the AAA sequence in the middle, you’d need to duplicate the A twice. But with only a single deletion, you wouldn’t be able to both fix the extra G at the start (for GG) and adjust the C at the end to match GGTAAAC.
• One duplication and two substitutions: In this case, duplicating A and substituting one T could form part of the AAA sequence, but you’d be left with only one substitution. That wouldn’t be enough to both create the GG at the beginning and ensure the sequence ends with C.

Since neither approach works within 3 steps, it’s clear that Option D cannot be completed in fewer than 4 mutations.

We are told that exactly one patient is ill, which leads us to three key pieces of information:

1. The infected individual must be one of patients 0 through 7, as indicated by the positive result from Test Tube A.
2. They cannot be patients 0 through 3, because Test Tube C confirmed all of them are healthy.
3. They also cannot be patients 4 or 5, since Test Tube E showed both of them are healthy.

With this information, the only possible patients who fit all three conditions are patients 6 and 7. The only test tubes that can differentiate between these two patients are Test Tubes G and H. Since Test Tube G is the only one listed as an option, it must be the correct answer. Thus, you should choose option D.

Let’s focus on the top left $2 \times 2$ section of the box. This section has $4$ compartments, and each one can either contain a marble or be empty. This means there are $$16\$$ different ways Hira can arrange marbles in this section. One possible arrangement might look like this:
Now, if we look at the first column, and it only has one marble, Hira will need to place a marble in compartment A to ensure the column total is even. On the other hand, if the second column already has an even number of marbles, Hira must leave compartment B empty to maintain that balance. Following the same reasoning, Hira would leave compartment C empty and place a marble in compartment D.

In general, once the marbles are placed in the top left $2 \times 2$ section, Hira has no choice when it comes to filling compartments A, B, C, and D. The requirement for each column and row to have an even total dictates whether each of these compartments must have a marble or not.

What about compartment E?

If the bottom row (not counting E) contains an odd number of marbles, Hira has two possibilities: either A has a marble and B is empty, or the reverse. Suppose A has a marble and B is empty. Because A has a marble, the two compartments directly above A must be different (one marble, one empty). Since B is empty, the two compartments above B must be the same (both marbles or both empty). This difference in pattern means that compartments C and D will be different from each other. As a result, the third column (without considering E) also contains an odd number of marbles. With both the bottom row and third column (excluding E) having odd totals, Hira is forced to place a marble in compartment E to make both totals even.

Similarly, if the bottom row (excluding E) contains an even number of marbles, the third column (again excluding E) will also contain an even number. In this case, Hira must leave compartment E empty to maintain the even totals.

In summary, no matter how Hira arranges the marbles in the top left $2 \times 2$ section, there is exactly one way to place marbles in the remaining five compartments in order to follow the even total rules. The only choice Hira really has is which of the 16 possible marble arrangements to use for the $2 \times 2$ section. You should choose option B.