Question

solve 5.3 5.8 by multiplying with whole numbers. solve 8 5 tens, and add 2 bundled tens. 5.3 * 5.8 ≈ 30 5 3 x 5 8 2 4

Explanation:

Step1: Analyze the multiplication (units place)

We are multiplying \(53\times58\) (since \(5.3\) and \(5.8\) are treated as \(53\) and \(58\) when multiplying as whole numbers, then we will adjust for decimals later, but here we focus on the multiplication steps). First, multiply the units digits: \(3\times8 = 24\). So the units place of the product is \(4\), and we carry over \(2\) (the tens digit of \(24\)).

Step2: Multiply units with tens (first part)

Now, multiply the units digit of the first number (\(3\)) with the tens digit of the second number (\(5\)): \(3\times5 = 15\). Then add the carried over \(2\): \(15 + 2=17\). Wait, no, actually, in the standard multiplication algorithm, we first do \(53\times8\) and \(53\times50\), then add. Wait, looking at the partial multiplication: the first partial product is \(53\times8\). Let's compute \(53\times8\): \(50\times8=400\), \(3\times8 = 24\), so \(400 + 24=424\). Wait, but in the given problem, the multiplication is set as \(53\times58\), and we are looking at the middle digit. Wait, maybe the problem is about the multiplication of \(53\times58\) and we need to find the middle digit. Wait, let's do \(53\times58\):

First, \(53\times8 = 424\) (this is the first partial product, with the last digit aligned with the units place). Then \(53\times50=2650\) (this is the second partial product, with the last digit aligned with the tens place). Now, add \(424+2650\):

\[ \begin{align*} &\quad\ 424\\ &+\ 2650\\ &\quad\ 3074 \end{align*} \]

But in the given problem, the multiplication is shown as:

\[ \begin{array}{c@{}c@{}c} & 5 & 3 \\ \times & 5 & 8 \\ \cline{1-3} & \square & \square & 4 \\ \end{array} \]

Wait, maybe the problem is focusing on the step where we multiply \(53\times8\) first. Wait, \(53\times8 = 424\), so the first partial product is \(424\), so the middle digit (tens place) of the first partial product is \(2\)? No, wait, the given numbers in the multiplication are \(53\times58\), and the first multiplication step (multiplying by \(8\)): \(53\times8 = 424\). So the digits of \(424\) are \(4\) (units), \(2\) (tens), \(4\) (hundreds)? Wait, no, \(424\) is \(4\) hundreds, \(2\) tens, and \(4\) units. Wait, maybe the problem is about the multiplication of \(53\times58\) and the boxed digits. Wait, looking at the image, the first partial product (from \(53\times8\)) is \(424\), but the way it's written, maybe we are doing the multiplication as:

First, multiply \(3\times8 = 24\) (units place \(4\), carry \(2\)). Then multiply \(5\times8 = 40\), add the carry \(2\): \(40 + 2=42\). Wait, that makes sense! Because \(53\times8=(50 + 3)\times8=50\times8+3\times8 = 400+24 = 424\). So breaking it down: \(3\times8 = 24\) (units: \(4\), carry \(2\)), then \(5\times8=40\), add carry \(2\) to get \(42\). So the first partial product is \(424\). So in the given format, the first two digits (after the carry) would be \(4\) and \(2\)? Wait, no, the image shows a multiplication like:

\[ \begin{array}{c@{}c@{}c} & \boldsymbol{5} & 3 \\ \times & & 8 \\ \cline{1-3} & \boldsymbol{4} & \boldsymbol{2} & 4 \\ \end{array} \]

Wait, maybe the problem is about the multiplication of \(53\times58\) and the middle digit when we do the first part (\(53\times8\)). Wait, in \(53\times8 = 424\), the tens digit is \(2\) and the hundreds digit is \(4\). But the image has a blue box on the first digit and a white box on the second digit, with the last digit \(4\). So the first blue box is \(4\) (hundreds place) and the white box is \(2\) (tens place)? Wait, no, maybe the problem is a typo or miswriting, but following the multiplication of \(53\times8 = 424\), the digits are \(4\) (hundreds), \(2\) (tens), \(4\) (units). So the middle digit (the white box) should be \(2\), and the first blue box is \(4\)? Wait, no, the first blue box is \(1\)? Wait, maybe I made a mistake. Wait, let's re - express the multiplication as \(53\times58\):

\[ \begin{array}{c@{}c@{}c} & 5 & 3 \\ \times & 5 & 8 \\ \cline{1-3} & 4 & 2 & 4 \quad (\text{from } 53\times8)\\ 2 & 6 & 5 & 0 \quad (\text{from } 53\times50)\\ \cline{1-4} 3 & 0 & 7 & 4 \end{array} \]

But the problem seems to be focusing on the first partial product (\(53\times8\)). So \(53\times8 = 424\), so the digits are \(4\) (hundreds), \(2\) (tens), \(4\) (units). So the white box (the middle digit) should be \(2\), and the first blue box (the hundreds digit) should be \(4\)? But the image shows a blue box with \(1\)? Wait, maybe the problem is about \(53\times58\) and there is a miscalculation, but following the multiplication of \(53\times8 = 424\), the middle digit is \(2\).

Answer:

The middle digit (the white box) is \(2\), and the first blue box (if we consider the first partial product \(53\times8 = 424\)) has the hundreds digit \(4\), but based on the multiplication steps for the units and tens part, the middle digit (tens place of the first partial product) is \(2\). So the answer for the white box is \(2\).