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7 4 6 2
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Question

name:
7 4 6 2
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 4 6
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  4 2
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  0
5 8 1 0
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3 6 6 0
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Accepted Answer

To solve these division problems, we follow the long - division algorithm. Let's take each problem one by one.

### First Problem (Completed as an Example)
We have $7462\div42$.
- First, we see how many times $42$ goes into $74$. $42\times1 = 42$, $42\times2=84>74$, so we take $1$. $74 - 42 = 32$.
- Bring down the $6$ to get $326$. Now, we find how many times $42$ goes into $326$. $42\times7 = 294$, $42\times8 = 336>326$, so we take $7$. $326-294 = 32$.
- Bring down the $2$ to get $322$. Now, we find how many times $42$ goes into $322$. $42\times7 = 294$, $42\times8 = 336>322$, so we take $7$. $322 - 294=28$? Wait, no, in the given example, the last step gives a remainder of $0$. Wait, maybe the number is $462\div42$ first? Wait, the first grid has $7$ on the left, and the numbers $4$, $6$, $2$ in the dividend part. Let's re - examine. The first problem seems to be a long - division of a number (maybe $462$) by $42$ (since the divisor steps involve $42$). $42\times11 = 462$, but the steps show subtracting $42$ multiple times. Anyway, let's move to the second and third problems.

### Second Problem: Dividend is $5810$, let's assume the divisor is related to the steps. Wait, looking at the first problem, the divisor might be a two - digit number. Let's look at the first problem's divisor steps: we have subtractions of $42$ (since $46 - 42=4$, then $42-42 = 0$). Wait, maybe the divisor is $42$ for the first, and for the second, let's see the dividend digits $8$, $1$, $0$ with a $5$ on the left? Wait, maybe the first number is $7462$ divided by $42$, the second is $5810$ divided by some number, and the third is $3660$ divided by some number.

Wait, maybe the pattern is that we are doing long division where we break down the dividend into parts. Let's take the second problem:

#### Step 1: Analyze the dividend and divisor
The dividend digits are $8$, $1$, $0$ (with a $5$ on the left, maybe the dividend is $5810$) and we need to find the divisor. Wait, looking at the first problem, the divisor is $42$ (since $42\times11 = 462$, but the first problem's dividend part is $4$, $6$, $2$ which is $462$). So $462\div42 = 11$. Let's check: $42\times10=420$, $462 - 420 = 42$, $42\div42 = 1$, so $10 + 1=11$.

For the second problem, the dividend part is $8$, $1$, $0$ (maybe $810$) and the left digit is $5$? Wait, no, maybe the dividend is $5810$ and the divisor is $5810\div x$. Wait, maybe the divisor is $5810\div10 = 581$, but that doesn't make sense. Alternatively, maybe the first problem is $462\div42 = 11$, the second is $810\div45 = 18$ (since $45\times18 = 810$), and the third is $660\div60 = 11$ (since $60\times11 = 660$). But this is getting confusing. Let's try to do the long division for the second problem:

Assume we are dividing $5810$ by $581$ (but that would be $10$). Or maybe the dividend is $810$ and the divisor is $45$. Let's do $810\div45$:

- Step 1: How many times does $45$ go into $81$? $45\times1 = 45$, $45\times2 = 90>81$, so we take $1$. $81-45 = 36$.
- Step 2: Bring down the $0$ to get $360$. How many times does $45$ go into $360$? $45\times8 = 360$. So $1\times10+8 = 18$. So $810\div45 = 18$.

For the third problem, the dividend digits are $6$, $6$, $0$ with a $3$ on the left (maybe the dividend is $3660$ or $660$). Let's assume the dividend is $660$ and the divisor is $60$:

- Step 1: How many times does $60$ go into $66$? $60\times1 = 60$, $60\times2 = 120>66$, so we take $1$. $66 - 60=6$.
- Step 2: Bring down the $0$ to get $60$. How many times does $60$ go into $60$? $1$ time. So $1\times10 + 1=11$. So $660\div60 = 11$.

But since the problem is about filling in the long - division steps, let's focus on the second problem (middle one):

#### Step 1: Divide the first part of the dividend
The first part of the dividend is $8$ (tens place?) Wait, maybe the dividend is $5810$ and we are dividing by $581$. But that's a three - digit divisor. Alternatively, let's look at the first problem's structure. The first problem has a divisor of $42$, as we saw $462\div42 = 11$.

For the second problem, let's assume the divisor is $581$ (since the left digit is $5$ and the dividend digits are $8$, $1$, $0$). Then $5810\div581$:

- Step 1: $581\times10 = 5810$, so the quotient is $10$.

For the third problem, the dividend is $3660$ and the divisor is $366$ (left digit $3$, dividend digits $6$, $6$, $0$):

- Step 1: $366\times10 = 3660$, so the quotient is $10$.

But this is a bit unclear. Alternatively, if we follow the first problem's pattern where we have a two - digit divisor:

First problem: $462\div42 = 11$ (since $42\times10 = 420$, $462 - 420 = 42$, $42\div42 = 1$, $10 + 1=11$)

Second problem: Let's take the dividend as $810$ and divisor as $45$ (since $45\times18 = 810$)

- Step 1: $45\times10 = 450$, $810 - 450 = 360$
- Step 2: $45\times8 = 360$, $360 - 360 = 0$
- Quotient: $10+8 = 18$

Third problem: Dividend $660$, divisor $60$ (since $60\times11 = 660$)

- Step 1: $60\times10 = 600$, $660 - 600 = 60$
- Step 2: $60\times1 = 60$, $60 - 60 = 0$
- Quotient: $10 + 1=11$

Since the problem is about long division, we can present the solutions as follows:

### Second Problem (Dividend $5810$ - assuming divisor $581$):
# Explanation:
## Step 1: Divide by 581
We know that $581\times10=5810$. So when we divide $5810$ by $581$, we have:
$5810\div581 = 10$

### Third Problem (Dividend $3660$ - assuming divisor $366$):
# Explanation:
## Step 1: Divide by 366
We know that $366\times10 = 3660$. So when we divide $3660$ by $366$, we have:
$3660\div366=10$

But if we follow the first problem's two - digit divisor pattern:

### Second Problem (Dividend $810$, Divisor $45$):
# Explanation:
## Step 1: Divide 81 by 45
$45\times1 = 45$, $81 - 45=36$
## Step 2: Bring down 0, get 360
$45\times8 = 360$, $360 - 360 = 0$
## Step 3: Find quotient
The quotient is $1 + 8=18$ (since we had $1$ in the tens place and $8$ in the ones place)
$810\div45 = 18$

### Third Problem (Dividend $660$, Divisor $60$):
# Explanation:
## Step 1: Divide 66 by 60
$60\times1 = 60$, $66 - 60 = 6$
## Step 2: Bring down 0, get 60
$60\times1 = 60$, $60 - 60 = 0$
## Step 3: Find quotient
The quotient is $1+1 = 11$ (since we had $1$ in the tens place and $1$ in the ones place)
$660\div60 = 11$

Since the first problem's solution ends with a remainder of $0$, we can assume that we are looking for exact divisions. The most probable answers based on the long - division structure are:

For the second problem (dividend $5810$ - if divisor is $581$): $\boldsymbol{10}$

For the third problem (dividend $3660$ - if divisor is $366$): $\boldsymbol{10}$

Or if we take two - digit divisors:

Second problem: $\boldsymbol{18}$

Third problem: $\boldsymbol{11}$

Given the first problem's steps, the two - digit divisor approach is more likely. So:

# Answer:
Second problem quotient: $\boldsymbol{18}$ (if dividend is $810$, divisor $45$)

Third problem quotient: $\boldsymbol{11}$ (if dividend is $660$, divisor $60$)

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QUESTION IMAGE

Question

Explanation:

First Problem (Completed as an Example)

Step 1: Analyze the dividend and divisor

Step 1: Divide by 581

Third Problem (Dividend \(3660\) - assuming divisor \(366\)):

Step 1: Divide by 366

Second Problem (Dividend \(810\), Divisor \(45\)):

Step 1: Divide 81 by 45

Step 2: Bring down 0, get 360

Step 3: Find quotient

Third Problem (Dividend \(660\), Divisor \(60\)):

Answer: