Let's solve the problem step-by-step.
We need the final digit of the product [tex]\(163 \times 87 \times \square \times 239\)[/tex] to be 1. Let's first examine the unit digit of the product [tex]\(163 \times 87 \times 239\)[/tex].
1. Determine the Last Digit of the Product [tex]\(163 \times 87 \times 239\)[/tex]:
- The unit digit of 163 is 3.
- The unit digit of 87 is 7.
- The unit digit of 239 is 9.
To find the unit digit of their product, we multiply their unit digits:
[tex]\[3 \times 7 = 21 \quad \text{(unit digit is 1)}\][/tex]
[tex]\[1 \times 9 = 9 \quad \text{(unit digit is 9)}\][/tex]
So, the unit digit of the product [tex]\(163 \times 87 \times 239\)[/tex] is 9.
2. Find the Unit Digit for [tex]\(\square\)[/tex] to Make the Final Product's Unit Digit 1:
- Let [tex]\(d\)[/tex] be the digit in place of [tex]\(\square\)[/tex].
- We need to find a digit [tex]\(d\)[/tex] such that when 9 (the unit digit of the intermediate product) multiplied by [tex]\(d\)[/tex], the unit digit of the resulting product is 1.
Let's check the products of 9 with digits 0 through 9 until we get a unit digit of 1:
[tex]\[
\begin{align*}
9 \times 0 & = 0 \quad (\text{unit digit is 0}) \\
9 \times 1 & = 9 \quad (\text{unit digit is 9}) \\
9 \times 2 & = 18 \quad (\text{unit digit is 8}) \\
9 \times 3 & = 27 \quad (\text{unit digit is 7}) \\
9 \times 4 & = 36 \quad (\text{unit digit is 6}) \\
9 \times 5 & = 45 \quad (\text{unit digit is 5}) \\
9 \times 6 & = 54 \quad (\text{unit digit is 4}) \\
9 \times 7 & = 63 \quad (\text{unit digit is 3}) \\
9 \times 8 & = 72 \quad (\text{unit digit is 2}) \\
9 \times 9 & = 81 \quad (\text{unit digit is 1})
\end{align*}
\][/tex]
Therefore, the digit [tex]\(d\)[/tex] that makes the unit digit of the product 1 is 9.
Conclusion: The digit that should be in place of [tex]\(\square\)[/tex] to make the last digit of the product [tex]\(163 \times 87 \times \square \times 239\)[/tex] be 1 is 9.
