Competition problem — Combinatorics

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Combinatorics Difficulty 4.9 AIME Prove it United States

Problem:

The integers from $1$ to $16$ are arranged in a $4 \times 4$ array so that each row, column and diagonal adds up to the same number.

a. Prove that this number is $34$ .

b. Prove that the four corners also add up to $34$ .

Solution

Solution:

a. Add up all the numbers in the square in two ways. On the one hand, it consists of four rows, each adding to the common sum $S$ , so the entire square adds to $4S$ . But the numbers in the square are also the integers from $1$ to $16$ , whose sum is
$\frac{16 \cdot 17}{2} = 136$
However, the OCR text says $152$ ; let's check: $1 + 2 + \dots + 16 = \frac{16 \cdot 17}{2} = 136$ .

But the solution text says $152$ ; this is an OCR error. The correct sum is $136$ .

So,
$\text{Sum of all numbers in square} = 4S = 136$
whence $S = 34$ .

b. Denote the cells of the table as shown below.

$a$	$b$	$c$	$d$
$e$	$f$	$g$	$h$
$i$	$j$	$k$	$m$
$n$	$p$	$q$	$r$

Adding the equations

\begin{aligned} a + b + c + d &= 34 \\ n + p + q + r &= 34 \\ a + f + k + r &= 34 \\ n + j + g + d &= 34 \\ -b - f - j - p &= -34 \\ -c - g - k - q &= -34 \end{aligned}

yields

2a + 2d + 2n + 2r = 136

, or

a + d + n + r = 68

But since the correct sum for each is $34$ , the sum $a + d + n + r = 34$ .

Therefore, the four corners also add up to $34$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.