An old book listed that 72 turkeys were bought for $\$\square.\square\square$ each, at a total cost of $\$\square 67.9\square.$
What was the cost of each turkey, if each $\square$ stands for a digit that was unreadable in the old book?
Note: The $\square$ digits are not all the same.
To solve this problem, we need to use a system of equations and algebraic manipulation to find the values of the unreadable digits.
Let's start by assigning variables to the unknown digits. Since there are three unreadable digits, we'll use $a$, $b$, and $c$ as placeholders.
Now let's break down the information given in the problem step by step.
The book says that 72 turkeys were bought for $\$\square.\square\square$ each. We can express the cost of each turkey as $10a + b + \frac{c}{100}$. Since the digits are unreadable, we need to multiply the first digit by 10 and the second digit by 1 to get their respective places.
The total cost of the turkeys is given as $\$\square 67.9\square$, which can be expressed as $10^3a + 10^2b + 67 + \frac{9}{10} + \frac{c}{100}$. Here, we have the 67 representing the tens and units digits, and 9/10 representing the decimal point.
Now we have two equations:
$72(10a + b + \frac{c}{100}) = 10^3a + 10^2b + 67 + \frac{9}{10} + \frac{c}{100}$
$720a + 72b + 0.72c = 1000a + 100b + 67.9 + c$
We can simplify this equation by converting the decimal numbers to fractions:
$720a + 72b + \frac{72c}{100} = 1000a + 100b + \frac{679}{10} + \frac{c}{100}$
Multiplying both sides of the equation by 100 to eliminate the fractions:
$72000a + 7200b + 72c = 100000a + 10000b + 6790 + c$
Now we have an equation without decimals:
$28000a + 2800b + 71c = 6790$
To find the values of $a$, $b$, and $c$, we need to solve this equation.
One way to solve this equation is through trial and error. Since $a$, $b$, and $c$ are digits, we know they must be integers between 0 and 9.
Start by substituting different values for $c$ (0 to 9) and solving for $a$ and $b$ using the equation:
$28000a + 2800b + 71c = 6790$
Continue substituting different values for $c$ until you find a combination of $a$, $b$, and $c$ that satisfies the equation. Remember that $a$, $b$, and $c$ cannot be the same, as stated in the problem.
For example, let's assume $c=5$. Plugging this into the equation, we have:
$28000a + 2800b + 71(5) = 6790$
$28000a + 2800b + 355 = 6790$
We can continue with this process until we find a solution that satisfies the equation.
I'll use x for ease of typing
each turkey cost x.xx/72
72*x.xx = x67.9x
start listing multiples of 72 that end in 9x:
72*11 = 792
72*18 = 1296
72*36 = 2592
Check to see which works out