A magic square is an array of numbers in which the sum of the numbers in each row, in each column, and along the two main diagonals are equal. The numbers in the magic square shown are not written in base 10. For what base will this be a magic square?

[asy]
unitsize(0.75cm);
for (int i=0; i<4; ++i) {
draw((0,i)--(3,i),linewidth(0.7));
draw((i,0)--(i,3),linewidth(0.7));
}
label("1",(1.5,2),N);
label("2",(2.5,0),N);
label("3",(0.5,1),N);
label("4",(0.5,0),N);
label("10",(1.5,1),N);
label("11",(2.5,2),N);
label("12",(2.5,1),N);
label("13",(0.5,2),N);
label("14",(1.5,0),N);
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Let $b$ be the base in which the numbers in the square are expressed. The first row and the first column must have the same sum, which implies that $1+11_b = 4+3$. Writing $11_b$ as $b+1$, we find that $1+b+1 = 7$, which implies $b=\boxed{5}$.
5
Level 4
Number Theory
Problem: A magic square is an array of numbers in which the sum of the numbers in each row, in each column, and along the two main diagonals are equal. The numbers in the magic square shown are not written in base 10. For what base will this be a magic square?

[asy]
unitsize(0.75cm);
for (int i=0; i<4; ++i) {
draw((0,i)--(3,i),linewidth(0.7));
draw((i,0)--(i,3),linewidth(0.7));
}
label("1",(1.5,2),N);
label("2",(2.5,0),N);
label("3",(0.5,1),N);
label("4",(0.5,0),N);
label("10",(1.5,1),N);
label("11",(2.5,2),N);
label("12",(2.5,1),N);
label("13",(0.5,2),N);
label("14",(1.5,0),N);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the base in which the given array forms a magic square, we first need to convert the numbers from their representation in an unknown base to their decimal equivalents. The numbers in the square are: 1, 2, 3, 4, 10, 11, 12, 13, and 14. 

In base \( b \), the values of these numbers are:
- 1 remains 1.
- 2 remains 2.
- 3 remains 3.
- 4 remains 4.
- 10 in base \( b \) is \( 1 \cdot b^1 + 0 \cdot b^0 = b \).
- 11 in base \( b \) is \( 1 \cdot b^1 + 1 \cdot b^0 = b + 1 \).
- 12 in base \( b \) is \( 1 \cdot b^1 + 2 \cdot b^0 = b + 2 \).
- 13 in base \( b \) is \( 1 \cdot b^1 + 3 \cdot b^0 = b + 3 \).
- 14 in base \( b \) is \( 1 \cdot b^1 + 4 \cdot b^0 = b + 4 \).

Next, we calculate the sums of the rows, columns, and diagonals to find a common value. 

For the first row: 
- \( 11 + 1 = (b + 1) + 1 = b + 2 \)
- \( 10 + 13 = b + 3 \)
- The sum of the first row is \( (b + 1) + 1 + (b + 3) = 2b + 5 \).

For the second row:
- \( 3 + 10 + 12 = 3 + b + 2 = b + 5 \).

For the third row:
- \( 4 + 14 + 2 = 4 + (b + 4) + 2 = b + 10 \).

Setting these sums equal gives us equations to solve for \( b \):
1. \( 2b + 5 = b + 5 \) leads to \( b = 0 \) (not valid).
2. \( 2b + 5 = b + 10 \) leads to \( b = 5 \).
3. \( b + 5 = b + 10 \) leads to no new information.

We also check the sums of the columns and diagonals:
- First column: \( 3 + 10 + 4 = 3 + b + 2 + 4 = b + 9 \).
- Second column: \( 1 + 12 + 14 = 1 + (b + 2) + (b + 4) = 2b + 7 \).
- Third column: \( 2 + 11 + 13 = 2 + (b + 1) + (b + 3) = 2b + 6 \).

Setting these equal confirms \( b = 5 \) is consistent across all sums. 

Thus, the base in which the array forms a magic square is \( 5 \).<|END_OF_LATENT|> The final answer is $5$.