1.9 division ii w
56
Division II http://www.lahc.edu/math/frankma.htm
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Transcript of 1.9 division ii w
Division IIn 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 back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
Division IIn 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 back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Division IIn 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 back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
Division IIn the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Division I
Enter the quotient on top
2
In the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
multiply the quotientback into the scaffold.
62 x 3 1
Division IIn the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
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.
multiply the quotientback into the scaffold.
62 x 3 1
Division IIn the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
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.
multiply the quotientback into the scaffold.
62 x 3 1
The new dividend is 1, not enough to be divided again, so stop. This is the remainder.
Division IIn the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
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.
multiply the quotientback into the scaffold.
62 x 3 1
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.
Division IIn the last section we demonstrated long division for 7 ÷ 3:
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).
Steps. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
“back-one” outside )3 7
“front-one” inside
Enter the quotient on top
2
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.
multiply the quotientback into the scaffold.
62 x 3 1
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.
Put the result in the multiplicative form, we have that 7 = 2 x 3 + 1.
Division IIn the last section we demonstrated long division for 7 ÷ 3:
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
ii. Place the quotient above the right end of the part of the dividend that is sufficient to be divided .
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
ii. Place the quotient above the right end of the part of the dividend that is sufficient to be divided
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
8
Enter and place the quotient above the right end of the part of the dividend.
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
8
Enter and place the quotient above the right end of the part of the dividend.
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
8multiply the quotient back
8 x 7
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.)7 5 9
8multiply the quotient back
8 x 7
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
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.
)7 5 98multiply the
quotient back8 x 7
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
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.
)7 5 98multiply the
quotient back8 x 7
The difference is 3, not enough to be divided again, stop.
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
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.
)7 5 98multiply the
quotient back8 x 7
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.
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Let’s refine this procedure for division of large numbers with the example 59 ÷ 7.
Division II
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.
Steps for long division. i. (Front-in Back-out)Put the problem in the long division format with the back-number (the divisor) outside, and the front-number (the dividend) inside the scaffold.
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.
)7 5 98multiply the
quotient back8 x 7
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.
3
Enter and place the quotient above the right end of the part of the dividend.
5 6
Restate the division result in the multiplicative form,it’s 59 = 7 x 8 + 3.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
Example A. Divide 598 ÷ 7.Find the quotient and the remainder.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8
Example A. Divide 598 ÷ 7.Find the quotient and the remainder.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
Example A. Divide 598 ÷ 7.Find the quotient and the remainder.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
Example A. Divide 598 ÷ 7.Find the quotient and the remainder.
But the two-left digits, or 59, is enough to be divided by 7.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
But the two-left digits, or 59, is enough to be divided by 7.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
iv. If the new dividend is sufficient to be divided by the divisor, repeat the process .
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
iv. If the new dividend is sufficient to be divided by the divisor, repeat the process .
5
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
iv. If the new dividend is sufficient to be divided by the divisor, repeat the process .
5
3 53
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
iv. If the new dividend is sufficient to be divided by the divisor, repeat the process .If not, this is the remainder.
5
3 53Remainder
For divisions of larger numbers, we first have to determine the placement of the quotient.
Division II
)7 5 9 8i. Starting from the left of the dividend, the digit 5 is not enough to be divided by 7.
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.
iii. Subtract the product of the quotient with the divisor, 8x7=56.
ii. So place the 1st quotient, which is 8, above the digit 9, i.e. above the right end of “59”.
8
But the two-left digits, or 59, is enough to be divided by 7.
5 6 3
Bring down the rest of the digits, this is the new dividend.
8
iv. If the new dividend is sufficient to be divided by the divisor, repeat the process .If not, this is the remainder.
5
3 53
Hence 598 ÷ 7 = 85 with R = 3or that 598 = 7 x 85 + 3.
Remainder
Division IIIt’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.
Division II
c. (Filling in 0’s) Divide 919 ÷ 9.
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.
)9 9 1 9
Division II
)9 9 1 91
9
c. (Filling in 0’s) Divide 919 ÷ 9.
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.
i. Starting from the left of the dividend, 9 goes into 9 once.
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
9iii. Bring down the rest of the digits, this is the new dividend.
c. (Filling in 0’s) Divide 919 ÷ 9.
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.
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9
c. (Filling in 0’s) Divide 919 ÷ 9.
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.
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9
c. (Filling in 0’s) Divide 919 ÷ 9.
0
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.
iv. 1 is not enough to be divided by 9, so the quotient is 0.
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9
2
c. (Filling in 0’s) Divide 919 ÷ 9.
0v. Enter the 2 as the quotient for 19 divided by 9.
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.
iv. 1 is not enough to be divided by 9, so the quotient is 0.
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9 vi. Continue the
process, subtract the product 2x9=18,
c. (Filling in 0’s) Divide 919 ÷ 9.
0
8
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.
iv. 1 is not enough to be divided by 9, so the quotient is 0.
1
v. Enter the 2 as the quotient for 19 divided by 9.
2
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9 vi. Continue the
process, subtract the product 2x9=18,we have R=1, stop.1
c. (Filling in 0’s) Divide 919 ÷ 9.
0
8
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.
iv. 1 is not enough to be divided by 9, so the quotient is 0.
1
v. Enter the 2 as the quotient for 19 divided by 9.
2
Division II
)9 9 1 9
i. Starting from the left of the dividend, 9 goes into 9 once.
ii. Subtract the product 1x9.
1
91
iii. Bring down the rest of the digits, this is the new dividend. 9 vi. Continue the
process, subtract the product 2x9=18,we have R=1, stop.1
Hence 919 ÷ 9 = 102 with remainder 1 or that 909 = 9 x 101.
c. (Filling in 0’s) Divide 919 ÷ 9.
0
8
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.
iv. 1 is not enough to be divided by 9, so the quotient is 0.
1
v. Enter the 2 as the quotient for 19 divided by 9.
2
Division II
)3 7 7 4 3 1 7
c. Divide 74317 ÷ 37. Find the Q and R.
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice. 2
c. Divide 74317 ÷ 37. Find the Q and R.
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
2
c. Divide 74317 ÷ 37. Find the Q and R.
7 4
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
2
c. Divide 74317 ÷ 37. Find the Q and R.
7 4
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
2
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
7 4
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
2
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
v. The two skipped-spaces must be filled by two “0’s”.
7 4
0 0
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
2
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
v. The two skipped-spaces must be filled by two “0’s”.
7 4
80 0
One checks that the quotient is 8.
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
vi. Continue, subtract 8x37=296
2
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
v. The two skipped-spaces must be filled by two “0’s”.
7 4
80 0
2 9 6
One checks that the quotient is 8.
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
vi. Continue, subtract 8x37=296 so R=21, which is not enough to be divided by 37, so stop.
2
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
v. The two skipped-spaces must be filled by two “0’s”.
7 4
80 0
2 9 62 1
One checks that the quotient is 8.
Division II
)3 7 7 4 3 1 7
i. Starting from the left,
37 goes into 74 twice.
ii. Subtract 2x37.
3 1 7iii. Bring down the rest of the digits, this is the new dividend.
vi. Continue, subtract 8x37=296 so R=21, which is not enough to be divided by 37, so stop.
2
Hence 74317 ÷ 37 = 2008 with R = 21,or that 74317 = 2008(37) + 21.
c. Divide 74317 ÷ 37. Find the Q and R.
iv. We need the entire 317 to be divided by 37.
v. The two skipped-spaces must be filled by two “0’s”.
7 4
80 0
2 9 62 1
One checks that the quotient is 8.
