CAT — Relative Ranking

From Condition (3), every ball except B3 pinged H1. Therefore, H1 is larger than B1, B2, B4, B5, and B6, while B3 is larger than H1.

Using Condition (4), only B2 pinged H2. Hence, B2 is larger than H2, whereas B1, B3, B4, B5, and B6 are all smaller than H2. This establishes that H2 is the smallest hoop.

From Condition (1), B1 and B6 pinged H4, but B5 did not. Therefore, B1 and B6 are larger than H4, while B5 is smaller than H4.

Similarly, from Condition (2), B4 pinged H3, whereas B1 did not. Hence, B4 is larger than H3, while B1 is smaller than H3.

Combining all these observations, we obtain the following order:

B2 < H2 < B4 < H3 < B1, B6 < H4 < B5 < H1 < B3

Accordingly, the only hoop order that satisfies all the conditions is:

H2 < H3 < H4 < H1

The deductions can be summarized in the table below.

The number of pings made by B1, B2, and B3 are 2, 4, and 0, respectively.

Hence, the required sum is: 2 + 4 + 0 = 6.

From Condition (3), every ball except B3 pinged H1. Therefore, H1 is larger than B1, B2, B4, B5, and B6, while B3 is larger than H1.

Using Condition (4), only B2 pinged H2. Hence, B2 is larger than H2, whereas B1, B3, B4, B5, and B6 are all smaller than H2. This establishes that H2 is the smallest hoop.

From Condition (1), B1 and B6 pinged H4, but B5 did not. Therefore, B1 and B6 are larger than H4, while B5 is smaller than H4.

Similarly, from Condition (2), B4 pinged H3, whereas B1 did not. Hence, B4 is larger than H3, while B1 is smaller than H3.

Combining all these observations, we obtain the following order:

B2 < H2 < B4 < H3 < B1, B6 < H4 < B5 < H1 < B3

Accordingly, the only hoop order that satisfies all the conditions is:

H2 < H3 < H4 < H1

The deductions can be summarized in the table below.

The number of pings made by B1, B2, and B3 are 2, 4, and 0, respectively.

There is no information, on the basis of which B6 and B1 can be compared. Hence, B1 < B6 < B3 is not necessarily true.

From Condition (3), every ball except B3 pinged H1. Therefore, H1 is larger than B1, B2, B4, B5, and B6, while B3 is larger than H1.

Using Condition (4), only B2 pinged H2. Hence, B2 is larger than H2, whereas B1, B3, B4, B5, and B6 are all smaller than H2. This establishes that H2 is the smallest hoop.

From Condition (1), B1 and B6 pinged H4, but B5 did not. Therefore, B1 and B6 are larger than H4, while B5 is smaller than H4.

Similarly, from Condition (2), B4 pinged H3, whereas B1 did not. Hence, B4 is larger than H3, while B1 is smaller than H3.

Combining all these observations, we obtain the following order:

B2 < H2 < B4 < H3 < B1, B6 < H4 < B5 < H1 < B3

Accordingly, the only hoop order that satisfies all the conditions is:

H2 < H3 < H4 < H1

The deductions can be summarized in the table below.

The number of pings made by B1, B2, and B3 are 2, 4, and 0, respectively.

We can see that H2 < H3 < H4 < H1 is true.

From Condition (3), every ball except B3 pinged H1. Therefore, H1 is larger than B1, B2, B4, B5, and B6, while B3 is larger than H1.

Using Condition (4), only B2 pinged H2. Hence, B2 is larger than H2, whereas B1, B3, B4, B5, and B6 are all smaller than H2. This establishes that H2 is the smallest hoop.

From Condition (1), B1 and B6 pinged H4, but B5 did not. Therefore, B1 and B6 are larger than H4, while B5 is smaller than H4.

Similarly, from Condition (2), B4 pinged H3, whereas B1 did not. Hence, B4 is larger than H3, while B1 is smaller than H3.

Combining all these observations, we obtain the following order:

B2 < H2 < B4 < H3 < B1, B6 < H4 < B5 < H1 < B3

Accordingly, the only hoop order that satisfies all the conditions is:

H2 < H3 < H4 < H1

The deductions can be summarized in the table below.

The number of pings made by B1, B2, and B3 are 2, 4, and 0, respectively.

We can see from the table that the total number of pings can be 12 or 13.

Step 1:

Only the values corresponding to B, C, E, and F are required to answer the questions.

In 2016, B had the highest SI, while F had the lowest. Also, the SI of B exceeded that of F by 60.

Let the SI of F in 2016 be x. Then, the SI of B in 2016 is x + 60.

Using the percentage changes shown in the graph, the SI of F becomes 2x in 2020 and 1.5x in 2024.

The SI of E in 2024 is 90. Applying the given percentage changes, the corresponding SI values of E are 75 in 2020 and 60 in 2016.

Since the range of SI values in 2024 is 60, the SI of F in 2024 (i.e., 1.5x) must be at least 30. Therefore, x ≥ 20, implying (x + 60) ≥ 80.

Now consider the percentage changes for B. Its SI decreases by 25% from 2016 to 2020, and again by 25% from 2020 to 2024. Thus, the overall multiplying factor from 2016 to 2024 is 9/16.

Since all SI values are integers, (x + 60) must be a multiple of 16. As it is already at least 80, the only possible values are 80 and 96.

• If (x + 60) = 96, then x = 36. This gives the SI of B in 2020 as 72, while the SI of F in 2020 also becomes 72, contradicting the fact that F had the lowest SI in 2020.
• Hence, this case is rejected.

Therefore, (x + 60) = 80, giving x = 20.

Thus, the SI values of B are:

• 2016: 80
• 2020: 60
• 2024: 45

Step 2:

In 2016, the SI of C lies between those of B and E. Therefore, its value must lie between 60 and 80.

From the graph, the SI of C becomes 4/5 of its 2016 value in 2020, and then 7/5 of its 2020 value in 2024. Hence, the overall multiplying factor from 2016 to 2024 is 28/25.

This implies that the SI of C in 2016 must be a multiple of 25. The only multiple of 25 between 60 and 80 is 75.

Accordingly, the SI of C is:

• 2016: 75
• 2020: 60
• 2024: 84

The derived values are summarized in the table below.

Hence, the SI of E in 2016 was 60.

Step 1:

Only the values corresponding to B, C, E, and F are required to answer the questions.

In 2016, B had the highest SI, while F had the lowest. Also, the SI of B exceeded that of F by 60.

Let the SI of F in 2016 be x. Then, the SI of B in 2016 is x + 60.

Using the percentage changes shown in the graph, the SI of F becomes 2x in 2020 and 1.5x in 2024.

The SI of E in 2024 is 90. Applying the given percentage changes, the corresponding SI values of E are 75 in 2020 and 60 in 2016.

Since the range of SI values in 2024 is 60, the SI of F in 2024 (i.e., 1.5x) must be at least 30. Therefore, x ≥ 20, implying (x + 60) ≥ 80.

Now consider the percentage changes for B. Its SI decreases by 25% from 2016 to 2020, and again by 25% from 2020 to 2024. Thus, the overall multiplying factor from 2016 to 2024 is 9/16.

Since all SI values are integers, (x + 60) must be a multiple of 16. As it is already at least 80, the only possible values are 80 and 96.

• If (x + 60) = 96, then x = 36. This gives the SI of B in 2020 as 72, while the SI of F in 2020 also becomes 72, contradicting the fact that F had the lowest SI in 2020.
• Hence, this case is rejected.

Therefore, (x + 60) = 80, giving x = 20.

Thus, the SI values of B are:

• 2016: 80
• 2020: 60
• 2024: 45

Step 2:

In 2016, the SI of C lies between those of B and E. Therefore, its value must lie between 60 and 80.

From the graph, the SI of C becomes 4/5 of its 2016 value in 2020, and then 7/5 of its 2020 value in 2024. Hence, the overall multiplying factor from 2016 to 2024 is 28/25.

This implies that the SI of C in 2016 must be a multiple of 25. The only multiple of 25 between 60 and 80 is 75.

Accordingly, the SI of C is:

• 2016: 75
• 2020: 60
• 2024: 84

The derived values are summarized in the table below.

The SI of F in 2020 = 40.

Step 1:

Only the values corresponding to B, C, E, and F are required to answer the questions.

In 2016, B had the highest SI, while F had the lowest. Also, the SI of B exceeded that of F by 60.

Let the SI of F in 2016 be x. Then, the SI of B in 2016 is x + 60.

Using the percentage changes shown in the graph, the SI of F becomes 2x in 2020 and 1.5x in 2024.

The SI of E in 2024 is 90. Applying the given percentage changes, the corresponding SI values of E are 75 in 2020 and 60 in 2016.

Since the range of SI values in 2024 is 60, the SI of F in 2024 (i.e., 1.5x) must be at least 30. Therefore, x ≥ 20, implying (x + 60) ≥ 80.

Now consider the percentage changes for B. Its SI decreases by 25% from 2016 to 2020, and again by 25% from 2020 to 2024. Thus, the overall multiplying factor from 2016 to 2024 is 9/16.

Since all SI values are integers, (x + 60) must be a multiple of 16. As it is already at least 80, the only possible values are 80 and 96.

• If (x + 60) = 96, then x = 36. This gives the SI of B in 2020 as 72, while the SI of F in 2020 also becomes 72, contradicting the fact that F had the lowest SI in 2020.
• Hence, this case is rejected.

Therefore, (x + 60) = 80, giving x = 20.

Thus, the SI values of B are:

• 2016: 80
• 2020: 60
• 2024: 45

Step 2:

In 2016, the SI of C lies between those of B and E. Therefore, its value must lie between 60 and 80.

From the graph, the SI of C becomes 4/5 of its 2016 value in 2020, and then 7/5 of its 2020 value in 2024. Hence, the overall multiplying factor from 2016 to 2024 is 28/25.

This implies that the SI of C in 2016 must be a multiple of 25. The only multiple of 25 between 60 and 80 is 75.

Accordingly, the SI of C is:

• 2016: 75
• 2020: 60
• 2024: 84

The derived values are summarized in the table below.

The SI of C in 2024 = 84

Step 1:

Only the values corresponding to B, C, E, and F are required to answer the questions.

In 2016, B had the highest SI, while F had the lowest. Also, the SI of B exceeded that of F by 60.

Let the SI of F in 2016 be x. Then, the SI of B in 2016 is x + 60.

Using the percentage changes shown in the graph, the SI of F becomes 2x in 2020 and 1.5x in 2024.

The SI of E in 2024 is 90. Applying the given percentage changes, the corresponding SI values of E are 75 in 2020 and 60 in 2016.

Since the range of SI values in 2024 is 60, the SI of F in 2024 (i.e., 1.5x) must be at least 30. Therefore, x ≥ 20, implying (x + 60) ≥ 80.

Now consider the percentage changes for B. Its SI decreases by 25% from 2016 to 2020, and again by 25% from 2020 to 2024. Thus, the overall multiplying factor from 2016 to 2024 is 9/16.

Since all SI values are integers, (x + 60) must be a multiple of 16. As it is already at least 80, the only possible values are 80 and 96.

• If (x + 60) = 96, then x = 36. This gives the SI of B in 2020 as 72, while the SI of F in 2020 also becomes 72, contradicting the fact that F had the lowest SI in 2020.
• Hence, this case is rejected.

Therefore, (x + 60) = 80, giving x = 20.

Thus, the SI values of B are:

• 2016: 80
• 2020: 60
• 2024: 45

Step 2:

In 2016, the SI of C lies between those of B and E. Therefore, its value must lie between 60 and 80.

From the graph, the SI of C becomes 4/5 of its 2016 value in 2020, and then 7/5 of its 2020 value in 2024. Hence, the overall multiplying factor from 2016 to 2024 is 28/25.

This implies that the SI of C in 2016 must be a multiple of 25. The only multiple of 25 between 60 and 80 is 75.

Accordingly, the SI of C is:

• 2016: 75
• 2020: 60
• 2024: 84

The derived values are summarized in the table below.

The SI of B in 2024 = 45

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

Hence, the PI of Whimshire is 45.

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Fogglia = 35

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Humbleset = 50

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The pair of cities that definitely belong to the same states Noodleton and Quackford.

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

We can identify the pIs of all cities and NURs and also identify the state they belong to.

Step 1:

Since 10 is the largest number, it must occupy the cell in Row 1, Column 4.

Also, 2 and 3 cannot lie in the same row or column as 10.

Now consider the placement of 9, the next largest number. It can be placed either in Row 1, Column 3 or Row 2, Column 4.

Step 2:

Assume 9 is placed in Row 2, Column 4.

In that case, 8 and 7 cannot share the same row or column with 9, forcing them into fixed positions.

However, this arrangement leads to a contradiction.

If 4 and 6 are placed in Row 2, there is no valid position left for 2 and 3.

On the other hand, if 4 and 6 are placed in Row 3, the entries in Column 3 will no longer be in decreasing order from top to bottom.

Hence, this case is not feasible.

Step 3:

Therefore, 9 must be placed in Row 1, Column 3.

Consequently, 8 and 7 cannot be in the same row or column as 9, which fixes their positions.

If 4 and 6 are placed in Row 2, there is no valid location available for 2 and 3.

Hence, 4 and 6 must be placed in Row 1.

The final arrangement is:

• Row 1: 4, 6, 9, 10
• Row 2: 2/3, 5, 8
• Row 3: 3/2, 7
• Row 4: 1

Step 1:

Since 10 is the largest number, it must occupy the cell in Row 1, Column 4.

Also, 2 and 3 cannot lie in the same row or column as 10.

Now consider the placement of 9, the next largest number. It can be placed either in Row 1, Column 3 or Row 2, Column 4.

Step 2:

Assume 9 is placed in Row 2, Column 4.

In that case, 8 and 7 cannot share the same row or column with 9, forcing them into fixed positions.

However, this arrangement leads to a contradiction.

If 4 and 6 are placed in Row 2, there is no valid position left for 2 and 3.

On the other hand, if 4 and 6 are placed in Row 3, the entries in Column 3 will no longer be in decreasing order from top to bottom.

Hence, this case is not feasible.

Step 3:

Therefore, 9 must be placed in Row 1, Column 3.

Consequently, 8 and 7 cannot be in the same row or column as 9, which fixes their positions.

If 4 and 6 are placed in Row 2, there is no valid location available for 2 and 3.

Hence, 4 and 6 must be placed in Row 1.

The final arrangement is:

• Row 1: 4, 6, 9, 10
• Row 2: 2/3, 5, 8
• Row 3: 3/2, 7
• Row 4: 1

Step 1:

Since 10 is the largest number, it must occupy the cell in Row 1, Column 4.

Also, 2 and 3 cannot lie in the same row or column as 10.

Now consider the placement of 9, the next largest number. It can be placed either in Row 1, Column 3 or Row 2, Column 4.

Step 2:

Assume 9 is placed in Row 2, Column 4.

In that case, 8 and 7 cannot share the same row or column with 9, forcing them into fixed positions.

However, this arrangement leads to a contradiction.

If 4 and 6 are placed in Row 2, there is no valid position left for 2 and 3.

On the other hand, if 4 and 6 are placed in Row 3, the entries in Column 3 will no longer be in decreasing order from top to bottom.

Hence, this case is not feasible.

Step 3:

Therefore, 9 must be placed in Row 1, Column 3.

Consequently, 8 and 7 cannot be in the same row or column as 9, which fixes their positions.

If 4 and 6 are placed in Row 2, there is no valid location available for 2 and 3.

Hence, 4 and 6 must be placed in Row 1.

The final arrangement is:

• Row 1: 4, 6, 9, 10
• Row 2: 2/3, 5, 8
• Row 3: 3/2, 7
• Row 4: 1

Step 1:

Since 10 is the largest number, it must occupy the cell in Row 1, Column 4.

Also, 2 and 3 cannot lie in the same row or column as 10.

Now consider the placement of 9, the next largest number. It can be placed either in Row 1, Column 3 or Row 2, Column 4.

Step 2:

Assume 9 is placed in Row 2, Column 4.

In that case, 8 and 7 cannot share the same row or column with 9, forcing them into fixed positions.

However, this arrangement leads to a contradiction.

If 4 and 6 are placed in Row 2, there is no valid position left for 2 and 3.

On the other hand, if 4 and 6 are placed in Row 3, the entries in Column 3 will no longer be in decreasing order from top to bottom.

Hence, this case is not feasible.

Step 3:

Therefore, 9 must be placed in Row 1, Column 3.

Consequently, 8 and 7 cannot be in the same row or column as 9, which fixes their positions.

If 4 and 6 are placed in Row 2, there is no valid location available for 2 and 3.

Hence, 4 and 6 must be placed in Row 1.

The final arrangement is:

• Row 1: 4, 6, 9, 10
• Row 2: 2/3, 5, 8
• Row 3: 3/2, 7
• Row 4: 1

Step 1:

Since 10 is the largest number, it must occupy the cell in Row 1, Column 4.

Also, 2 and 3 cannot lie in the same row or column as 10.

Now consider the placement of 9, the next largest number. It can be placed either in Row 1, Column 3 or Row 2, Column 4.

Step 2:

Assume 9 is placed in Row 2, Column 4.

In that case, 8 and 7 cannot share the same row or column with 9, forcing them into fixed positions.

However, this arrangement leads to a contradiction.

If 4 and 6 are placed in Row 2, there is no valid position left for 2 and 3.

On the other hand, if 4 and 6 are placed in Row 3, the entries in Column 3 will no longer be in decreasing order from top to bottom.

Hence, this case is not feasible.

Step 3:

Therefore, 9 must be placed in Row 1, Column 3.

Consequently, 8 and 7 cannot be in the same row or column as 9, which fixes their positions.

If 4 and 6 are placed in Row 2, there is no valid location available for 2 and 3.

Hence, 4 and 6 must be placed in Row 1.

The final arrangement is:

• Row 1: 4, 6, 9, 10
• Row 2: 2/3, 5, 8
• Row 3: 3/2, 7
• Row 4: 1