Library / /25 of 66

, 2025

Algebra Difficulty 2.1 Junior Prove it Canada

In a magic square, the numbers in each of the three
rows, each of the three columns, and each of the two diagonals have the
same sum. This sum is called the magic constant. Each of the
four figures shown below is a magic square.

In Figure 1, the magic constant is $18$ . What is the value of $n$ ? Figure 1 $7$ $2$ $n$ $3$ In Figure 2, what is the value of $p$ ? Figure 2 $8$ $p$ $9$ $5$ $4$ In Figure 3, what is the value of $r$ ? Figure 3 $13$ $r$ $7$ $17$ $\!r\!+\!1\!$ $\!r\!+\!3\!$ In Figure 4, determine the value of $u$ . Figure 4 $\!u\!+\!3\!$ $12$ $\!u\!+\!2\!$ $\!u\!-\!5\!$ $u$

Solution

For each part, the completed magic square is shown following part
(d).

$7$ $2$ $n$ $3$ The magic constant is $18$ , and so the missing number in the first row is $18-7-2=9$ . Looking at the diagonal from the top-right corner to the bottom-left corner, we get $9+n+3=18$ , and so $n=6$ . $8$ $p$ $9$ $5$ $4$ Reading from the first column, the magic constant is $8+9+4=21$ . Thus, the missing number in the second row is $21-9-5=7$ . Looking at the diagonal from the top-right corner to the bottom-left corner, the missing number in the top-right corner is $21-7-4=10$ . From the first row, we get $8+p+10=21$ , and so $p=3$ . $13$ $r$ $7$ $17$ $\!r\!+\!1\!$ $\!r\!+\!3\!$ Solution 1: The sum of the numbers in the first column is equal to the sum of the numbers in the third row. Since these two sums both share the missing number in the bottom-left corner, then the sum of the remaining two numbers in the first column must equal the sum of the remaining two numbers in the third row. That is, $13+7=(r+1)+(r+3)$ and so $20=2r+4$ or $16=2r$ , which gives $r=8$ . Solution 2: The sum of the numbers in the third column is $r+17+(r+3)=2r+20$ , and so the sum of the numbers in the third row is also $2r+20$ . Thus, the missing number in the third row is $(2r+20)-(r+1)-(r+3)=16$ . From the first column, the magic constant is $13+7+16=36$ , and so $2r+20=36$ or $2r=16$ , which gives $r=8$ . $\!u\!+\!3\!$ $12$ $\!u\!+\!2\!$ $\!u\!-\!5\!$ $u$ The sum of the numbers in the third row is $(u+2)+(u-5)+u=3u-3$ , and so the sum of the numbers in the second column is also $3u-3$ . Thus, the missing number in the second column is $(3u-3)-(u+3)-(u-5)=u-1$ , as shown. $\!u\!+\!3\!$ $\!u\!-\!1\!$ $12$ $\!u\!+\!2\!$ $\!u\!-\!5\!$ $u$ The sum of the numbers in the diagonal from the top-right corner to the bottom-left corner is equal to the sum of the numbers in the third column. Since these two sums both share the missing number in the top-right corner, then the sum of the remaining two numbers in the diagonal must equal the sum of the remaining two numbers in the third column. That is, $(u+2)+(u-1)=u+12$ or $2u+1=u+12$ , and so $u=11$ . (a) $7$ $2$ $9$ $8$ $6$ $4$ $3$ $10$ $5$ (b) $8$ $3$ $10$ $9$ $7$ $5$ $4$ $11$ $6$ (c) $13$ $15$ $8$ $7$ $12$ $17$ $16$ $9$ $11$ (d) $9$ $14$ $7$ $8$ $10$ $12$ $13$ $6$ $11$

Figure for this problem

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.