1998 CEMC Gauss (Grade 8) Problems/Problem 12

Problem

In the $4\text{x}4$ square shown, each row, column, and diagonal should contain each of the numbers $1$ , $2$ , $3$ , and $4$ . Find the value of $K + N$ .

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$\text{ (A) }\ 4 \qquad\text{ (B) }\ 3 \qquad\text{ (C) }\ 5 \qquad\text{ (D) }\ 6 \qquad\text{ (E) }\ 7$

Solution 1

$R$ must be equal to $4$ because $1$ , $2$ , $3$ are already present in a diagonal that contains $R$ .

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$J$ must be $2$ or $4$ because $1$ and $3$ are in a column with $1$ , $3$ , $J$ , and $G$ . However, $2$ already appears in a row containing $J$ , so $J$ must be $4$ .

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We can also conclude $Q = 3$ , $P = 2$ , $T = 3$ , and $L = 4$ using similar logic.

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$3$ , $2$ , and $4$ are already present in the row with $K$ , so $K = 1$ .

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Since $N$ can't be $1$ due to $K$ being in the same column and being $1$ , $M$ = $1$ , which also shows $N = 2$ using similar logic to $K = 1$ .

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Therefore, $K + N$ = $2 + 1 = \boxed {\textbf {(B) } 3}$ .

We can also verify that our answer is correct by filling up all of the square:

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~anabel.disher

Solution 2 (shorter)

We can start out by getting the value of $R$ and $J$ , like in solution 1.

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However, we can see that $K = 1$ because $T$ or $K$ must be $1$ , but $T$ cannot be $1$ since it is in a column with $1$ .

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From the column with $N$ and $K$ , $N = 3$ or $N = 2$ because $1$ and $4$ are in the column already. However, $N$ cannot be $3$ because it is in a row with $3$ , so $N = 2$ .

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Therefore, $K + N$ = $2 + 1 = \boxed {\textbf {(B) } 3}$ .

~anabel.disher

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1998 CEMC Gauss (Grade 8) Problems/Problem 12

Problem

Solution 1

Solution 2 (shorter)