2024 AMC 10A Problem 21

P21-25~7 minPrint

Sequences & Series Systems of Equations Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Substitute to simplify (u = x+1/x, shifting, scaling)

The numbers, in order, of each row and the numbers, in order, of each column of a $5 \times 5$ array of integers form an arithmetic progression of length $5$ . The numbers in positions $(5, 5)$ , $(2, 4)$ , $(4, 3)$ and $(3, 1)$ are $0$ , $48$ , $16$ , and $12$ , respectively. What number is in position $(1, 2)$ ? $\begin{bmatrix} . & ? &.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ .&.&.&.&0\end{bmatrix}$

Show solution without answering

Key insight

Measure from the bottom-right 0. Column 5 goes up in equal steps, and the row steps themselves change by equal amounts, so each entry is P·u + v·(Q + R·u).

Solution

Four numbers are given, so we look for a formula for the grid with four unknowns. The corner $(5, 5)$ is $0$ , so it is easiest to measure from there.

1. New labels. For the entry in row $i$ , column $j$ , let $u = 5 - i$ (how many rows above the bottom) and $v = 5 - j$ (how many columns left of the right edge). The $0$ is at $u = 0$ , $v = 0$ .
2. The right-hand column. Column $5$ is arithmetic and starts at $0$ at the bottom. So going up it reads $0, P, 2P, 3P, 4P$ for some step $P$ : the entry at height $u$ is $Pu$ .
3. The row steps. Each row is arithmetic, with its own step going left; call the step in row $u$ by the name $s_u$ . Then $s_u$ is (column $4$ entry) minus (column $5$ entry) in that row. Both columns are arithmetic going up, so their difference also changes by the same amount each row. That means the row steps form an arithmetic sequence too: $s_u = Q + Ru$ for some numbers $Q$ and $R$ .
4. The formula. Start at the column-5 entry $Pu$ and move $v$ steps left:
$\text{entry} = Pu + v\,(Q + Ru) = Pu + Qv + Ruv.$
5. Use the three other given numbers.
- Position $(2, 4)$ : $u = 3$ , $v = 1$ . So $3P + Q + 3R = 48$ .
- Position $(4, 3)$ : $u = 1$ , $v = 2$ . So $P + 2Q + 2R = 16$ .
- Position $(3, 1)$ : $u = 2$ , $v = 4$ . So $2P + 4Q + 8R = 12$ ; halve it: $P + 2Q + 4R = 6$ .
6. Solve. The last two equations differ only in the $R$ term: subtracting gives $2R = 6 - 16 = -10$ , so $R = -5$ .
Then the second equation becomes $P + 2Q = 26$ , and the first becomes $3P + Q = 63$ .
From $P = 26 - 2Q$ : $3(26 - 2Q) + Q = 63$ , so $78 - 5Q = 63$ , $Q = 3$ , and $P = 20$ .
7. Check the three given numbers: $60 + 3 - 15 = 48$ , $20 + 6 - 10 = 16$ , $40 + 12 - 40 = 12$ . All correct.
8. Position $(1, 2)$ : $u = 4$ , $v = 3$ :
$20 \cdot 4 + 3 \cdot 3 + (-5) \cdot 4 \cdot 3 = 80 + 9 - 60 = 29.$

The answer is $\boxed{\textbf{(C)}\ 29}$ .

Why this works

When every row and every column is arithmetic, the whole grid is fixed by just four numbers: the steps can change from row to row, but only by a steady amount. Writing the entry as a formula with four unknowns and plugging in the four given entries does the rest. Putting the starting point at the given $0$ removes one unknown before you begin.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $x(i,j)$ be the entry in row $i$ , column $j$ . Row $i$ is arithmetic, so $x(i,j) = x(i,5) + (j-5)\,d_i$ where $d_i$ is that row's common difference. Column $5$ is arithmetic, so $x(i,5)$ is a linear function of $i$ . Also $x(i,4) - x(i,5) = -d_i$ is the difference of two columns, each arithmetic in $i$ , so $d_i$ is linear in $i$ as well. Therefore every entry has the form
$x(i,j) = \alpha + \beta i + \gamma j + \delta\, ij.$

Measure from the corner: set $u = 5 - i$ , $v = 5 - j$ . Since $x(5,5) = 0$ the constant term vanishes and $x = Pu + Qv + Ruv$ . The three other data points give

$(2,4)$ , i.e. $u = 3, v = 1$ : $3P + Q + 3R = 48$ ,
$(4,3)$ , i.e. $u = 1, v = 2$ : $P + 2Q + 2R = 16$ ,
$(3,1)$ , i.e. $u = 2, v = 4$ : $2P + 4Q + 8R = 12$ , or $P + 2Q + 4R = 6$ .

Subtracting the second from the third: $2R = -10$ , so $R = -5$ . Then $P + 2Q = 26$ and $3P + Q = 63$ ; from the first $P = 26 - 2Q$ , so $78 - 5Q = 63$ , giving $Q = 3$ and $P = 20$ .

Position $(1,2)$ has $u = 4$ , $v = 3$ :
$x = 20\cdot 4 + 3\cdot 3 + (-5)(4)(3) = 80 + 9 - 60 = 29.$

The answer is $\boxed{\textbf{(C)}\ 29}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 11: Special Factorizations
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)