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Algebra Difficulty 2.6 Junior Find the answer Canada

In a magic square, the numbers in each row, the numbers in each column, and the numbers on each diagonal have the same sum.
$\begin{array}{|c|c|c|} \hline a & 13 & b \\ \hline 19 & c & 11 \\ \hline 12 & d & 16 \\ \hline \end{array}$
In the magic square shown, the sum $a+b+c$ equals

Pick one

Solutions — 2

Solution 1

In a magic square, the numbers in each row, the numbers in each column, and numbers on each diagonal have the same sum.

Since the sum of the numbers in the first row equals the sum of the numbers in the first column, then $a+13+b=a+19+12$ or $b=19+12-13=18$ .

Therefore, the sum of the numbers in any row, in any column, or along either diagonal equals the sum of the numbers in the third column, which is $18+11+16=45$ .

Using the first column, $a+19+12=45$ or $a=14$ .

Using the second row, $19+c+11=45$ or $c=15$ .

Thus, $a+b+c=14+18+15=47$ .

Solution 2

In a magic square, the numbers in each row, the numbers in each column, and numbers on each diagonal have the same sum.

Since the sum of the numbers in the first row equals the sum of the numbers in the first column, then $a+13+b=a+19+12$ or $b=19+12-13=18$ .

Therefore, the sum of the numbers in any row, in any column, or along either diagonal equals the sum of the numbers in the third column, which is $18+11+16=45$ .

Using the first column, $a+19+12=45$ or $a=14$ .

Using the second row, $19+c+11=45$ or $c=15$ .

Thus, $a+b+c=14+18+15=47$ .

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