2. Division I
In the last section we demonstrated long division for 7 ÷ 3:
3. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
4. Division I
In the last section we demonstrated long division for 7 ÷ 3:
“back-one”
outside
“front-one”
inside
3 ) 7
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
5. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
“front-one”
inside
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
6. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
“front-one”
inside
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
7. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
6
2x3
1
multiply the quotient
back into the scaffold.
“front-one”
inside
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
8. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
6
2x3
1
multiply the quotient
back into the scaffold.
“front-one”
inside
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder.
Otherwise, repeat steps i and ii.
9. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
6
2x3
1
“front-one”
inside
multiply the quotient
back into the scaffold.
The new dividend is 1, not
enough to be divided again, so
stop. This is the remainder.
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder.
Otherwise, repeat steps i and ii.
10. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
6
2x3
1
“front-one”
inside
multiply the quotient
back into the scaffold.
The new dividend is 1, not
enough to be divided again, so
stop. This is the remainder.
So the remainder is 1 and
we have that 7 ÷ 3 = 2 with R = 1.
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder.
Otherwise, repeat steps i and ii.
11. Division I
In the last section we demonstrated long division for 7 ÷ 3:
Enter the
quotient on top
“back-one”
outside
2
3 ) 7
6
2x3
1
“front-one”
inside
multiply the quotient
back into the scaffold.
The new dividend is 1, not
enough to be divided again, so
stop. This is the remainder.
So the remainder is 1 and
we have that 7 ÷ 3 = 2 with R = 1.
Steps. i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Enter the quotient on top,
Multiply the quotient back into the
problem and subtract the results
from the dividend (and bring down
the rest of the digits, if any. This is
the new dividend).
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder.
Otherwise, repeat steps i and ii.
Put the result in the multiplicative form, we have that
7 = 2 x 3 + 1.
13. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
7)5 9
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
14. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
7)5 9
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided .
15. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
8
7)5 9
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided
16. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
8
7)5 9
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
17. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
18. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
Bring down the rest of the digits,
if any, to form the new dividend.
19. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
Bring down the rest of the digits,
if any, to form the new dividend.
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder
R. Otherwise, repeat steps i and ii.
20. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
The difference is 3, not enough
to be divided again, stop.
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
Bring down the rest of the digits,
if any, to form the new dividend.
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder
R. Otherwise, repeat steps i and ii.
21. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
The difference is 3, not enough
to be divided again, stop.
So the remainder is 3 and
we’ve that 59 ÷ 7 = 8 with R = 3.
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
Bring down the rest of the digits,
if any, to form the new dividend.
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder
R. Otherwise, repeat steps i and ii.
22. Division II
Let’s refine this procedure for division of large numbers with
the example 59 ÷ 7.
Steps for long division.
Enter and place the quotient
above the right end of the part
of the dividend.
multiply the
quotient back
8x7
8
7)5 9
56
3
The difference is 3, not enough
to be divided again, stop.
So the remainder is 3 and
we’ve that 59 ÷ 7 = 8 with R = 3.
Restate the division result in the
multiplicative form,
it’s 59 = 7 x 8 + 3.
i. (Front-in Back-out)
Put the problem in the long
division format with the backnumber (the divisor) outside, and
the front-number (the dividend)
inside the scaffold.
ii. Place the quotient above the
right end of the part of the
dividend that is sufficient to be
divided . Multiply the quotient back
into the problem and subtract the
results from the dividend.
Bring down the rest of the digits,
if any, to form the new dividend.
iii. If the new dividend is not
enough to be divided by the
divisor, stop. This is the remainder
R. Otherwise, repeat steps i and ii.
23. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient.
24. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
25. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
7) 5
9
8
26. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
7) 5
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
9
8
27. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
7) 5
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
9
8
28. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
7) 5
9
8
29. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
8
9
8
30. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
8
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
31. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
8
8
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
32. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
iv. If the new dividend is
sufficient to be divided by the
divisor, repeat the process .
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
8
8
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
33. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
iv. If the new dividend is
sufficient to be divided by the
divisor, repeat the process .
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
5
8
8
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
34. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
iv. If the new dividend is
sufficient to be divided by the
divisor, repeat the process .
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
3
5
8
8
5
3
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
35. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
iv. If the new dividend is
sufficient to be divided by the
divisor, repeat the process .
If not, this is the remainder.
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
3
Remainder
5
8
8
5
3
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
36. Division II
For divisions of larger numbers, we first have to determine the
placement of the quotient. Example A. Divide 598 ÷ 7.
Find the quotient and the remainder.
The first digit of the
quotient is to be placed
at the right end of the
part of the dividend that
is sufficient to be divided
by the divisor.
i. Starting from the left of the
dividend, the digit 5 is not
enough to be divided by 7.
But the two-left digits, or 59,
is enough to be divided by 7.
iv. If the new dividend is
sufficient to be divided by the
divisor, repeat the process .
If not, this is the remainder.
ii. So place the 1st quotient,
which is 8, above the digit 9,
i.e. above the right end of “59”.
7) 5
5
8
9
6
3
3
Remainder
5
8
8
5
3
iii. Subtract the
product of the
quotient with the
divisor, 8x7=56.
Bring down the
rest of the digits,
this is the new
dividend.
Hence 598 ÷ 7 = 85 with R = 3
or that 598 = 7 x 85 + 3.
37. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
38. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
9 )9
1
9
39. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
i. Starting from the left of the
dividend, 9 goes into 9 once.
1
9 )9 1
9
9
40. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1
9 )9 1
9
9
41. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1
9 )9 1
9
1
9
9
42. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
iv. 1 is not enough to
be divided by 9,
so the quotient is 0.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1 0
9 )9 1
9
1
9
9
43. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
iv. 1 is not enough to
be divided by 9,
so the quotient is 0.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1 0
9 )9 1
9
1
2
9
9
v. Enter the 2 as the
quotient for 19 divided
by 9.
44. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
iv. 1 is not enough to
be divided by 9,
so the quotient is 0.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1 0
9 )9 1
9
1
1
2
9
9
8
v. Enter the 2 as the
quotient for 19 divided
by 9.
vi. Continue the
process, subtract the
product 2x9=18,
45. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
iv. 1 is not enough to
be divided by 9,
so the quotient is 0.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1 0
9 )9 1
9
1
1
2
9
9
8
1
v. Enter the 2 as the
quotient for 19 divided
by 9.
vi. Continue the
process, subtract the
product 2x9=18,
we have R=1, stop.
46. Division II
It’s possible that when entering the new quotient there is one or
more spaces between it and the previous quotient, we must fill
in 0’s in those spaces.
c. (Filling in 0’s) Divide 919 ÷ 9.
iv. 1 is not enough to
be divided by 9,
so the quotient is 0.
i. Starting from the left of the
dividend, 9 goes into 9 once.
ii. Subtract the
product 1x9.
iii. Bring down the rest of
the digits, this is the new
dividend.
1 0
9 )9 1
9
1
1
2
9
9
8
1
v. Enter the 2 as the
quotient for 19 divided
by 9.
vi. Continue the
process, subtract the
product 2x9=18,
we have R=1, stop.
Hence 919 ÷ 9 = 102 with remainder 1 or that
909 = 9 x 101.
47. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
3 7 )7
4
3 1
7
48. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
i. Starting from the left,
37 goes into 74 twice.
3 7 )7
2
4
3 1
7
49. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 7 )7
7
2
4
4
3 1
7
50. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
3 7 )7
7
2
4
4
3 1
7
3 1
7
51. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
iv. We need the entire 317
to be divided by 37.
3 7 )7
7
2
4
4
3 1
7
3 1
7
52. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
v. The two skipped-spaces
must be filled by two “0’s”.
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
iv. We need the entire 317
to be divided by 37.
3 7 )7
7
2 0 0
4 3 1
4
3 1
7
7
53. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
v. The two skipped-spaces
must be filled by two “0’s”.
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
iv. We need the entire 317
to be divided by 37.
One checks that
the quotient is 8.
3 7 )7
7
2 0 0 8
4 3 1 7
4
3 1 7
54. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
v. The two skipped-spaces
must be filled by two “0’s”.
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
iv. We need the entire 317
to be divided by 37.
One checks that
the quotient is 8.
3 7 )7
7
2 0
4 3
4
3
2
0 8
1 7
1
9
7
6
vi. Continue, subtract
8x37=296
55. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
v. The two skipped-spaces
must be filled by two “0’s”.
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
iv. We need the entire 317
to be divided by 37.
One checks that
the quotient is 8.
3 7 )7
7
2 0
4 3
4
3
2
0 8
1 7
1
9
2
7
6
1
vi. Continue, subtract
8x37=296 so R=21,
which is not enough to
be divided by 37, so stop.
56. c. Divide 74317 ÷ 37.
Find the Q and R.
Division II
v. The two skipped-spaces
must be filled by two “0’s”.
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
iii. Bring down the rest of
the digits, this is the new
dividend.
3 7 )7
7
2 0
4 3
4
3
2
iv. We need the entire 317
to be divided by 37.
One checks that
the quotient is 8.
Hence 74317 ÷ 37 = 2008 with R = 21,
or that 74317 = 2008(37) + 21.
0 8
1 7
1
9
2
7
6
1
vi. Continue, subtract
8x37=296 so R=21,
which is not enough to
be divided by 37, so stop.
