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Difference between revisions of "2013 Mock AIME I Problems/Problem 4"

m (category)
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== Solution ==
 
== Solution ==
−
<math>\boxed{342}</math>.
+
 
 +
Let <math>P</math> be the "common product" of the magic square (i.e. the product of each row, diagonal, and column). There are now two cases: <math>P \neq 0</math> and <math>P=0</math>.
 +
 
 +
<math>\underline{\text{Case 1: } P \neq 0.}</math> Let the top right corner be <math>n</math>, as in the picture below. Then, we know that <math>P=9n</math>, so the middle left cell must have a value of <math>3n</math>:
 +
 
 +
<asy>
 +
size(100);
 +
defaultpen(linewidth(0.7));
 +
int i;
 +
for(i=0; i<4; i=i+1) {
 +
draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6));
 +
}
 +
label("$1$", (1,5));
 +
label("$9$", (3,5));
 +
label("$3$", (1,1));
 +
label("$n$", (5,5));
 +
label("$3n$", (1,3));
 +
</asy>
 +
 
 +
Looking at the diagonal which stretches from the bottom left to the top right, we see that the central cell must be <math>3</math>. Then, by looking at the other diagonal, we see that the bottom right cell must be <math>3n</math>:
 +
 
 +
<asy>
 +
size(100);
 +
defaultpen(linewidth(0.7));
 +
int i;
 +
for(i=0; i<4; i=i+1) {
 +
draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6));
 +
}
 +
label("$1$", (1,5));
 +
label("$9$", (3,5));
 +
label("$3$", (1,1));
 +
label("$n$", (5,5));
 +
label("$3n$", (1,3));
 +
label("$3$", (3,3));
 +
label("$3n$", (5,1));
 +
</asy>
 +
 
 +
Looking at the bottom row, we see that the bottom central cell has a value of <math>1</math>, and looking at the middle row tells us that the right middle cell also has a value of <math>1</math>:
 +
 
 +
<asy>
 +
size(100);
 +
defaultpen(linewidth(0.7));
 +
int i;
 +
for(i=0; i<4; i=i+1) {
 +
draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6));
 +
}
 +
label("$1$", (1,5));
 +
label("$9$", (3,5));
 +
label("$3$", (1,1));
 +
label("$n$", (5,5));
 +
label("$3n$", (1,3));
 +
label("$3$", (3,3));
 +
label("$3n$", (5,1));
 +
label("$1$", (3,1));
 +
label("$1$", (5,3));
 +
</asy>
 +
 
 +
Looking at the central column, we see that <math>9 \cdot 3 \cdot 1 = P = 9n</math>, so <math>n=3</math>. Because <math>3n=9 \leq 10</math> and all the rows, diagonals, and columns multiply to <math>27</math>, <math>n=3</math> satisfies the constriants of the problem. Thus, when <math>P \neq 0</math>, we have only <math>1</math> possibility for the magic square.
 +
 
 +
<math>\underline{\text{Case 2: }P=0.}</math> Because only one cell remains on both the left column and the top row, their remaining cells must be <math>0</math>, as shown below:
 +
 
 +
<asy>
 +
size(100);
 +
defaultpen(linewidth(0.7));
 +
int i;
 +
for(i=0; i<4; i=i+1) {
 +
draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6));
 +
}
 +
label("$1$", (1,5));
 +
label("$9$", (3,5));
 +
label("$3$", (1,1));
 +
label("$0$", (5,5));
 +
label("$0$", (1,3));
 +
</asy>
 +
 
 +
Now, we still need to place zeroes in the top left to bottom right diagonal, the middle column, and the bottom row. Placing a zero in the right middle cell will bring us no closer to this goal, so it can be any integer inclusively between <math>0</math> and <math>10</math>, and thus it can take <math>11</math> possible states. Let the three remaining cells be called "special cells." Placing two zeroes in any of the special cells will ensure that <math>P=0</math>, so we have another free cell which can also take <math>11</math> different values.
 +
 
 +
To prevent overcounting, we now have two cases. In the first case, '''only''' two of the special cells are zero, and so we have <math>3 \choose 2</math> <math>\cdot 10 \cdot 11 = 330</math> possibilities, because the third special cell cannot be zero. In the second case, all three special cells are zero, so we have only the <math>11</math> possibilities for the middle right cell. Adding these two numbers together, we get <math>330+11=341</math>.
 +
 
 +
Remembering to add the <math>1</math> possibility from Case 1, we get <math>341+1=\boxed{342}</math> as our final answer.
  
 
== See also ==
 
== See also ==

Revision as of 08:53, 30 July 2024

Problem

Compute the number of ways to fill in the following magic square such that:

1. the product of all rows, columns, and diagonals are equal (the sum condition is waived),

2. all entries are nonnegative integers less than or equal to ten, and

3. entries CAN repeat in a column, row, or diagonal.

[asy] size(100); defaultpen(linewidth(0.7)); int i; for(i=0; i<4; i=i+1) { draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6)); } label("$1$", (1,5)); label("$9$", (3,5)); label("$3$", (1,1)); [/asy]

Solution

Let $P$ be the "common product" of the magic square (i.e. the product of each row, diagonal, and column). There are now two cases: $P \neq 0$ and $P=0$.

$\underline{\text{Case 1: } P \neq 0.}$ Let the top right corner be $n$, as in the picture below. Then, we know that $P=9n$, so the middle left cell must have a value of $3n$:

[asy] size(100); defaultpen(linewidth(0.7)); int i; for(i=0; i<4; i=i+1) { draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6)); } label("$1$", (1,5)); label("$9$", (3,5)); label("$3$", (1,1)); label("$n$", (5,5)); label("$3n$", (1,3)); [/asy]

Looking at the diagonal which stretches from the bottom left to the top right, we see that the central cell must be $3$. Then, by looking at the other diagonal, we see that the bottom right cell must be $3n$:

[asy] size(100); defaultpen(linewidth(0.7)); int i; for(i=0; i<4; i=i+1) { draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6)); } label("$1$", (1,5)); label("$9$", (3,5)); label("$3$", (1,1)); label("$n$", (5,5)); label("$3n$", (1,3)); label("$3$", (3,3)); label("$3n$", (5,1)); [/asy]

Looking at the bottom row, we see that the bottom central cell has a value of $1$, and looking at the middle row tells us that the right middle cell also has a value of $1$:

[asy] size(100); defaultpen(linewidth(0.7)); int i; for(i=0; i<4; i=i+1) { draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6)); } label("$1$", (1,5)); label("$9$", (3,5)); label("$3$", (1,1)); label("$n$", (5,5)); label("$3n$", (1,3)); label("$3$", (3,3)); label("$3n$", (5,1)); label("$1$", (3,1)); label("$1$", (5,3)); [/asy]

Looking at the central column, we see that $9 \cdot 3 \cdot 1 = P = 9n$, so $n=3$. Because $3n=9 \leq 10$ and all the rows, diagonals, and columns multiply to $27$, $n=3$ satisfies the constriants of the problem. Thus, when $P \neq 0$, we have only $1$ possibility for the magic square.

$\underline{\text{Case 2: }P=0.}$ Because only one cell remains on both the left column and the top row, their remaining cells must be $0$, as shown below:

[asy] size(100); defaultpen(linewidth(0.7)); int i; for(i=0; i<4; i=i+1) { draw((0,2*i)--(6,2*i)^^(2*i,0)--(2*i,6)); } label("$1$", (1,5)); label("$9$", (3,5)); label("$3$", (1,1)); label("$0$", (5,5)); label("$0$", (1,3)); [/asy]

Now, we still need to place zeroes in the top left to bottom right diagonal, the middle column, and the bottom row. Placing a zero in the right middle cell will bring us no closer to this goal, so it can be any integer inclusively between $0$ and $10$, and thus it can take $11$ possible states. Let the three remaining cells be called "special cells." Placing two zeroes in any of the special cells will ensure that $P=0$, so we have another free cell which can also take $11$ different values.

To prevent overcounting, we now have two cases. In the first case, only two of the special cells are zero, and so we have $3 \choose 2$ $\cdot 10 \cdot 11 = 330$ possibilities, because the third special cell cannot be zero. In the second case, all three special cells are zero, so we have only the $11$ possibilities for the middle right cell. Adding these two numbers together, we get $330+11=341$.

Remembering to add the $1$ possibility from Case 1, we get $341+1=\boxed{342}$ as our final answer.

See also

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