3. In which number is the value of the first 8 ten times as great as the value of the second 8 ? (A) 82,810 (B) 38,183 (C) 28,831 (D) 28,581

Let the first \(8\) be in the place with value \(10^k\) and the second \(8\) be in the place with value \(10^m\). Their contributions are: \[ 8 \times 10^k \quad \text{and} \quad 8 \times 10^m \] We are told that the value of the first \(8\) is ten times the value of the second \(8\): \[ 8 \times 10^k = 10 \times \bigl(8 \times 10^m\bigr) \] Dividing both sides by \(8\) gives: \[ 10^k = 10 \times 10^m = 10^{m+1} \] Thus, we have: \[ k = m + 1 \] This means that the first \(8\) must be one place to the left of the second \(8\) (their place values differ by a factor of \(10\)). Now, examine the options: - **(A) \(82,\!810\):** Its digits are \(8, 2, 8, 1, 0\). The first \(8\) is in the ten-thousands place (\(8 \times 10^4 = 80,\!000\)) and the second \(8\) is in the hundreds place (\(8 \times 10^2 = 800\)). Their ratio is: \[ \frac{80,\!000}{800} = 100 \quad (\text{not } 10) \] - **(B) \(38,\!183\):** Its digits are \(3, 8, 1, 8, 3\). The first \(8\) is in the thousands place (\(8 \times 10^3 = 8,\!000\)) and the second \(8\) is in the tens place (\(8 \times 10^1 = 80\)). Their ratio is: \[ \frac{8,\!000}{80} = 100 \quad (\text{not } 10) \] - **(C) \(28,\!831\):** Its digits are \(2, 8, 8, 3, 1\). The first \(8\) is in the thousands place (\(8 \times 10^3 = 8,\!000\)) and the second \(8\) is in the hundreds place (\(8 \times 10^2 = 800\)). Their ratio is: \[ \frac{8,\!000}{800} = 10 \] - **(D) \(28,\!581\):** Its digits are \(2, 8, 5, 8, 1\). The first \(8\) is in the thousands place (\(8 \times 10^3 = 8,\!000\)) and the second \(8\) is in the tens place (\(8 \times 10^1 = 80\)). Their ratio is: \[ \frac{8,\!000}{80} = 100 \quad (\text{not } 10) \] Only option **(C) \(28,\!831\)** satisfies the condition.