The numbers 1 to 16 are written in the cells of a 4 × 4 grid, one number per cell. Rows are numbered 1 to 4 from top to bottom and columns 1 to 4 from left to right; the cell in row r and column c is written (r, c).
The numbers are placed so that for every n from 1 to 15, the cells containing n and n + 1 share a side (they are horizontally or vertically adjacent). In other words, the numbers 1, 2, …, 16 trace a single path through the grid that visits every cell exactly once, moving one step up, down, left or right each time.
It is further known that:
1. The number 1 is in cell (1, 1).
2. The number 2 is in cell (1, 2).
3. The sum of the four numbers in each row is the same for all four rows.
The numbers are placed so that for every n from 1 to 15, the cells containing n and n + 1 share a side (they are horizontally or vertically adjacent). In other words, the numbers 1, 2, …, 16 trace a single path through the grid that visits every cell exactly once, moving one step up, down, left or right each time.
It is further known that:
1. The number 1 is in cell (1, 1).
2. The number 2 is in cell (1, 2).
3. The sum of the four numbers in each row is the same for all four rows.
Question 1
Numerical answer
Which number is in cell (3, 2)?
Show answer and solution
Answer: 8
Solution. 1 + 2 + … + 16 = 136, so each row sums to 136/4 = 34. Row 1 holds 1 and 2, so (1, 3) and (1, 4) sum to 31 ⇒ they hold 15 and 16. 3 must touch 2, so 3 is at (2, 2).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Cell (3, 2) contains 8.
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Cell (3, 2) contains 8.
Question 2
Single correct option
What is the sum of the numbers on the diagonal from (1, 1) to (4, 4)?
- A. 24
- B. 34
- C. 30
- D. 44
Show answer and solution
Answer: (A) 24
Solution. 1 + 2 + … + 16 = 136, so each row sums to 136/4 = 34. Row 1 holds 1 and 2, so (1, 3) and (1, 4) sum to 31 ⇒ they hold 15 and 16. 3 must touch 2, so 3 is at (2, 2).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Diagonal: (1, 1) = 1, (2, 2) = 3, (3, 3) = 9, (4, 4) = 11. Sum = 24. (44 is the other diagonal: 16 + 14 + 8 + 6.)
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Diagonal: (1, 1) = 1, (2, 2) = 3, (3, 3) = 9, (4, 4) = 11. Sum = 24. (44 is the other diagonal: 16 + 14 + 8 + 6.)
Question 3
Single correct option
In which cell is the number 16?
- A. (1, 3)
- B. (1, 4)
- C. (4, 4)
- D. (4, 1)
Show answer and solution
Answer: (B) (1, 4)
Solution. 1 + 2 + … + 16 = 136, so each row sums to 136/4 = 34. Row 1 holds 1 and 2, so (1, 3) and (1, 4) sum to 31 ⇒ they hold 15 and 16. 3 must touch 2, so 3 is at (2, 2).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
16 is in the top-right corner, (1, 4).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
16 is in the top-right corner, (1, 4).
Question 4
Single correct option
What is the sum of the numbers in column 3?
- A. 42
- B. 46
- C. 48
- D. 50
Show answer and solution
Answer: (C) 48
Solution. 1 + 2 + … + 16 = 136, so each row sums to 136/4 = 34. Row 1 holds 1 and 2, so (1, 3) and (1, 4) sum to 31 ⇒ they hold 15 and 16. 3 must touch 2, so 3 is at (2, 2).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Column 3: 15 + 14 + 9 + 10 = 48.
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
Column 3: 15 + 14 + 9 + 10 = 48.
Question 5
Numerical answer
What is the sum of all the numbers in the cells that share a side with the cell containing 9?
Show answer and solution
Answer: 44
Solution. 1 + 2 + … + 16 = 136, so each row sums to 136/4 = 34. Row 1 holds 1 and 2, so (1, 3) and (1, 4) sum to 31 ⇒ they hold 15 and 16. 3 must touch 2, so 3 is at (2, 2).
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
9 is at (3, 3). Its neighbours: (2, 3) = 14, (4, 3) = 10, (3, 2) = 8, (3, 4) = 12. Sum = 14 + 10 + 8 + 12 = 44.
Key idea: every number other than 1 and 16 needs two path-neighbours (n − 1 and n + 1). Cell (2, 1) touches only 1, 3 and (3, 1), so it must hold 4 (it cannot be 2). Hence (2, 1) = 4 and 5 is at (3, 1).
*Case 16 at (1, 3), 15 at (1, 4):* 14 is at (2, 4), so (2, 3) = 34 − 4 − 3 − 14 = 13 and 12 is at (3, 3). Cell (3, 4) touches 14, 12 and (4, 4), so it must be 11 with 10 at (4, 4); row 3 then forces (3, 2) = 34 − 5 − 12 − 11 = 6, so 7 is at (4, 2). Now 8 and 9 go in (4, 1) and (4, 3), but 9 must touch both 8 and 10 – impossible. Rejected.
*Case 15 at (1, 3), 16 at (1, 4):* 14 is at (2, 3), so (2, 4) = 34 − 4 − 3 − 14 = 13, and 12 is at (3, 4). Row 3: (3, 2) + (3, 3) = 34 − 5 − 12 = 17. If 6 were at (3, 2), then (3, 3) = 11 and 10 at (4, 3), leaving 7, 8, 9 for (4, 1), (4, 2), (4, 4) with no valid path. So 6 is at (4, 1), 7 at (4, 2), and then 8 at (3, 2), 9 at (3, 3) (8 + 9 = 17 ✓), 10 at (4, 3), 11 at (4, 4).
| | c1 | c2 | c3 | c4 |
|---|---|---|---|---|
| r1 | 1 | 2 | 15 | 16 |
| r2 | 4 | 3 | 14 | 13 |
| r3 | 5 | 8 | 9 | 12 |
| r4 | 6 | 7 | 10 | 11 |
Every row sums to 34 and consecutive numbers are adjacent; an exhaustive check of all 552 possible paths confirms this is the only grid.
9 is at (3, 3). Its neighbours: (2, 3) = 14, (4, 3) = 10, (3, 2) = 8, (3, 4) = 12. Sum = 14 + 10 + 8 + 12 = 44.
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