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Difference between revisions of "1998 CEMC Gauss (Grade 8) Problems/Problem 12"

(Created page with "==Problem== In the <math>4\text{x}4</math> square shown, each row, column, and diagonal should contain each of the numbers <math>1</math>, <math>2</math>, <math>3</math>, an...")
 
 
Line 17: Line 17:
 
Therefore, <math>K + N</math> = <math>2 + 1 = \boxed {\textbf {(B) } 3}</math>.
 
Therefore, <math>K + N</math> = <math>2 + 1 = \boxed {\textbf {(B) } 3}</math>.
  
 +
We can also verify that our answer is correct by filling up all of the square:
 +
{{Image needed}}
 
~anabel.disher
 
~anabel.disher
−
==Solution 2==
+
==Solution 2 (shorter)==
 
We can start out by getting the value of <math>R</math> and <math>J</math>, like in solution 1.
 
We can start out by getting the value of <math>R</math> and <math>J</math>, like in solution 1.
 
{{Image needed}}
 
{{Image needed}}

Latest revision as of 15:03, 18 June 2025

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 $K = 1$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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:

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

~anabel.disher

Solution 2 (shorter)

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $1$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $N = 2$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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

~anabel.disher

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