A non-square rectangle has integer dimensions. The number  of square units in its area is numerically equal to the number of units in its perimeter. What is the number of units in the perimeter of this rectangle?
Let the two sides of the rectangle be $a$ and $b$.  The problem is now telling us $ab=2a+2b$.  Putting everything on one side of the equation, we have $ab-2a-2b=0.$  This looks tricky.  However, we can add a number to both sides of the equation to make it factor nicely.  4 works here:  $$ab-2a-2b+4=4 \Rightarrow (a-2)(b-2)=4$$Since we don't have a square, $a$ and $b$ must be different.  It doesn't matter which one is which, so we can just say $a=6 $ and $b=3 $.  The perimeter is then $2(6+3)=\boxed{18}$
18
Level 4
Algebra
Problem: A non-square rectangle has integer dimensions. The number  of square units in its area is numerically equal to the number of units in its perimeter. What is the number of units in the perimeter of this rectangle?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the length and width of the rectangle be denoted as L and W, respectively. The area A of the rectangle can be expressed as A = L * W. The perimeter P of the rectangle is given by P = 2L + 2W. 

According to the problem, the area is equal to the perimeter, leading to the equation:

L * W = 2L + 2W.

Rearranging this equation gives:

L * W - 2L - 2W = 0.

This can be factored or rearranged to find integer solutions. To facilitate this, we can rewrite the equation as:

L * W - 2L - 2W + 4 = 4.

This can be factored as:

(L - 2)(W - 2) = 4.

Now, we need to find pairs of integers (L - 2) and (W - 2) that multiply to 4. The integer factor pairs of 4 are (1, 4), (2, 2), and their negatives. However, since L and W must be positive integers, we only consider the positive pairs:

1. (1, 4) gives L - 2 = 1 and W - 2 = 4, leading to L = 3 and W = 6.
2. (4, 1) gives L - 2 = 4 and W - 2 = 1, leading to L = 6 and W = 3.
3. (2, 2) gives L - 2 = 2 and W - 2 = 2, leading to L = 4 and W = 4, but this is a square, which is not allowed.

Now we calculate the perimeter for the valid pairs:

For L = 3 and W = 6:
P = 2(3) + 2(6) = 6 + 12 = 18.

For L = 6 and W = 3:
P = 2(6) + 2(3) = 12 + 6 = 18.

Both configurations yield the same perimeter. Therefore, the only valid integer perimeter for the rectangle, given the conditions of the problem, is 18.<|END_OF_LATENT|> The final answer is $18$.