Answer :
We are given the division problem
[tex]$$
62 \div 5537.
$$[/tex]
We will solve it step by step, as follows.
1. \textbf{First Partial Division:}
Consider the first three digits of the dividend (553). We need to determine how many times [tex]$62$[/tex] can be subtracted from [tex]$553$[/tex].
Compute:
[tex]$$
62 \times 8 = 496 \quad\text{and}\quad 62 \times 9 = 558.
$$[/tex]
Since [tex]$558$[/tex] is too large (i.e. [tex]$558 > 553$[/tex]), we use [tex]$8$[/tex]. Thus, the first digit of the quotient is [tex]$8$[/tex].
2. \textbf{First Multiplication and Subtraction:}
Multiply [tex]$8$[/tex] by [tex]$62$[/tex]:
[tex]$$
8 \times 62 = 496.
$$[/tex]
Subtract this from [tex]$553$[/tex]:
[tex]$$
553 - 496 = 57.
$$[/tex]
3. \textbf{Bring Down the Next Digit:}
Bring down the next digit of the dividend (which is [tex]$7$[/tex]) next to the [tex]$57$[/tex] to form [tex]$577$[/tex].
4. \textbf{Second Partial Division:}
Now, determine how many times [tex]$62$[/tex] goes into [tex]$577$[/tex].
Compute:
[tex]$$
62 \times 9 = 558 \quad\text{and}\quad 62 \times 10 = 620.
$$[/tex]
Since [tex]$620 > 577$[/tex], we choose [tex]$9$[/tex]. Hence, the second digit of the quotient is [tex]$9$[/tex].
5. \textbf{Second Multiplication and Final Subtraction:}
Multiply [tex]$9$[/tex] by [tex]$62$[/tex]:
[tex]$$
9 \times 62 = 558.
$$[/tex]
Subtract this product from [tex]$577$[/tex]:
[tex]$$
577 - 558 = 19.
$$[/tex]
This [tex]$19$[/tex] is the final remainder.
6. \textbf{Final Answer:}
The overall quotient, obtained by combining the digits from the two steps, is:
[tex]$$
89.
$$[/tex]
The missing numbers (or the key values in the long division) are:
- The first product: [tex]$496$[/tex],
- The result after the first subtraction: [tex]$57$[/tex],
- The second product: [tex]$558$[/tex],
- And the final remainder: [tex]$19$[/tex].
Thus, the division of [tex]$5537$[/tex] by [tex]$62$[/tex] results in a quotient of [tex]$89$[/tex] with a remainder of [tex]$19$[/tex].
[tex]$$
62 \div 5537.
$$[/tex]
We will solve it step by step, as follows.
1. \textbf{First Partial Division:}
Consider the first three digits of the dividend (553). We need to determine how many times [tex]$62$[/tex] can be subtracted from [tex]$553$[/tex].
Compute:
[tex]$$
62 \times 8 = 496 \quad\text{and}\quad 62 \times 9 = 558.
$$[/tex]
Since [tex]$558$[/tex] is too large (i.e. [tex]$558 > 553$[/tex]), we use [tex]$8$[/tex]. Thus, the first digit of the quotient is [tex]$8$[/tex].
2. \textbf{First Multiplication and Subtraction:}
Multiply [tex]$8$[/tex] by [tex]$62$[/tex]:
[tex]$$
8 \times 62 = 496.
$$[/tex]
Subtract this from [tex]$553$[/tex]:
[tex]$$
553 - 496 = 57.
$$[/tex]
3. \textbf{Bring Down the Next Digit:}
Bring down the next digit of the dividend (which is [tex]$7$[/tex]) next to the [tex]$57$[/tex] to form [tex]$577$[/tex].
4. \textbf{Second Partial Division:}
Now, determine how many times [tex]$62$[/tex] goes into [tex]$577$[/tex].
Compute:
[tex]$$
62 \times 9 = 558 \quad\text{and}\quad 62 \times 10 = 620.
$$[/tex]
Since [tex]$620 > 577$[/tex], we choose [tex]$9$[/tex]. Hence, the second digit of the quotient is [tex]$9$[/tex].
5. \textbf{Second Multiplication and Final Subtraction:}
Multiply [tex]$9$[/tex] by [tex]$62$[/tex]:
[tex]$$
9 \times 62 = 558.
$$[/tex]
Subtract this product from [tex]$577$[/tex]:
[tex]$$
577 - 558 = 19.
$$[/tex]
This [tex]$19$[/tex] is the final remainder.
6. \textbf{Final Answer:}
The overall quotient, obtained by combining the digits from the two steps, is:
[tex]$$
89.
$$[/tex]
The missing numbers (or the key values in the long division) are:
- The first product: [tex]$496$[/tex],
- The result after the first subtraction: [tex]$57$[/tex],
- The second product: [tex]$558$[/tex],
- And the final remainder: [tex]$19$[/tex].
Thus, the division of [tex]$5537$[/tex] by [tex]$62$[/tex] results in a quotient of [tex]$89$[/tex] with a remainder of [tex]$19$[/tex].
