Long Division

Report
How To Use MrV’s Times Table
- For Long Division!
1. Introduction:
a. Long Division – Terms and Concepts
b. Find the Position of the 1st Quotient Digit
2. Division by a 1-Digit Divisor:
a.
b.
c.
d.
e.
Divide a 2-Digit Dividend by a 1-Digit Divisor
When a Quotient has a Remainder
When a Dividend is Past the last Product
When a Division has More than One Step
When a Quotient has More Than Two Digits
3. Division by Multiple-Digit Divisors
a. Using “Round & Simplify” to Estimate Quotient Digits
b. When a Quotient has an Internal or Ending Zero
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Long Division Terms & Concepts
• Key words:
Quotient
Divisor Dividend
• For this division example,
Click to show
–
–
–
–
–
–
–
The Divisor
The Dividend
The Quotient
The 1st Quotient Digit
The Partial Dividends
The Products to be Subtracted
The Remainder
• Click again when done
Row 7 from MrV’s Times Table:
Find the Position of the 1st Quotient Digit
• Terms:
Quotient
Divisor Dividend
“_” is called “underline.”
A “partial dividend” is used
to find a single quotient digit.
• Build the Quotient one digit at a time, starting from the left-most position.
• How do you find the position of the 1st Quotient Digit?
• Critical skill: Putting the first quotient digit in the correct position
(underlining the 1st digit position is recommended, but it’s not required. Don’t use red )
4?
28
?7?
5 35
?4???
?3
???
??1
?? ?
3 123 22 6688 402 66885
• Bonus: You will know exactly how many digits will be in the quotient
(the quotient may or may not have a remainder)
Divide a 2-digit Dividend by a 1-digit Divisor
(when one number divides evenly into another)
1. Find the row
for the divisor
8_
7 56
 56
0
Please notice:
Each table cell has a
product with its
two factors below it
3. Use the 2nd
factor under the
product as the
quotient digit
2. Look across the
row until you find
a product equal to
the dividend
Prove it! Subtract the product from the dividend to get 0 (no remainder)
Ok, try a few even divisions, step by step:
??5
8 40
 40
0
7
??
4 28
 28
0
8
??
9 72
 72
0
??6
6 36
 36
0
1. Find 1st quotient digit’s place. 2. Find the row. 3. Find the product. 4. The 2nd factor is the value. 5. Prove it!
When a Quotient has a Remainder
(when one number does not divide evenly into another)
1. Find the row
for the divisor
5
8 43
 40
3r 3
43
2. Look across the
row until you see
the last product
that is less than
the dividend
3. Use the 2nd
factor under the
product as the
quotient digit
4. Subtract the product from the dividend to get the remainder;
Show the remainder with a small r just to the right of the quotient
When a Dividend is Past the last Product
3 28
 27
1r1
28
(there will be a Remainder)
9
1. Find the row
for the divisor
3. The quotient digit
is the second factor
under the product
2. Stop at the last
product in the row
if it is still less than
the dividend
4. Subtract the product from the dividend to get the remainder;
Show the remainder with a small r just to the right of the quotient
Ok, let’s try a few divisions with remainders:
r2
8
5 42
 40
2
r3
9
4 39
 36
3
42
39
r5
5
7 40
 35
5
r8
3
9 35
 27
8
1. Find the row and the product. 2. The 2nd factor is the digit 3. Subtract the product to find the remainder
40
35
When Division has More Than One Step
4 85
8
05
4
1r1
(for each step, only a partial dividend is used)
4. The 1st quotient
digit is the 2nd factor
under the product
5
1. Find the position of
the 1st quotient digit
The 1st quotient digit is 2
Subtract product 8 from 8,
leaving 0
1 2
5. Create the 2nd
partial dividend by
bringing down the next
digit. Use the table to
find the product and
the 2nd quotient digit.
In this case, it’s above the 8,
making 8 the 1st partial dividend
The quotient will be 2 digits long
2. Find the row
for the divisor
Bring down 5, making 05.
Ignore any leading 0’s:
5 is the 2nd partial dividend
5 is right after product 4,
so 1 is the 2nd quotient digit.
In this case, it’s row 4
3. Stop at the last
product that does not
exceed the partial
dividend (it can be =)
In this case 8 is equal to product 8
6. Subtract the product from the dividend to get the remainder;
Show the remainder with a small r just to the right of the quotient
5 – 4 = 1, which is the remainder
Another 2-Digit Quotient Example
(You still only need one row from MrV’s Times Table – show work under the division)
1. Find the position of
the first quotient digit:
5 does not go into 3, but it will divide
into 35 (the 1st partial dividend) – so
the 1st quotient digit goes above the 5
2. Find the value of the
first quotient digit, and
put it in its place:
Use MrV’s Table, Row 5:
35 exactly matches product 35,
so the 1st quotient digit is the factor 7
3. Subtract the product from
the dividend. Then bring down
the next dividend digit next to
the subtraction number
Subtract 35 from 35, getting 0
Bring down 0: The last partial
dividend becomes 00
You are now ready to use 0 to find
the second quotient digit
?7?_0?
5 350
 35
00
 0
0
5. Repeat steps 2 and 3 for the last
digit in the dividend:
In MrV’s Table, 0 is exactly on the 0 product,
so the factor 0 is the last quotient digit
Subtract 0 from 00, getting 0
Since there is a 0 remainder, we are finished
When a quotient has more than 1 digit, you
will need to break the dividend into smaller
pieces (call them partial dividends).
When a Quotient has More Than Two Digits
(You only need one row from MrV’s Times Table – show work under the division)
1. Find the position of
the first quotient digit:
7 does not go into 3, but it will divide
into 39 – so the first quotient digit
has to go right above the 9.
2. Find the value of the
first quotient digit, and
put it in its place:
Use MrV’s Table, Row 7:
39 is between products 35 and 42,
so the first quotient digit is 5
3. Subtract the product from
the dividend. Then bring down
the next dividend digit next to
the subtraction number
Subtract 35 from 39, getting 4
Bring down 3: 4 becomes 43
You are now ready to use 43 as the
2nd partial dividend to find the
second quotient digit
? 5?,
?8?
_,6?1
7 3939,328
 35
4 343
 42
1 2 12
 7
58 58
 56
2 r2
5. Repeat steps 2 and 3 for each
remaining digit in the dividend:
In MrV’s Table, 43 is between products 42 and 47,
so 6 is the 2nd quotient digit
Subtract 42 from 43, getting 1
Bring down 2: 1 becomes 12 (3rd partial dividend)
In MrV’s Table, 12 is between products 7 and 14,
so 1 is the 3rd quotient digit
Subtract 7 from 12, getting 5
Bring down 8: 5 becomes 58 (4th partial dividend)
In MrV’s Table, 58 is between products 56 and 63,
so 8 is the 4th quotient digit
Subtract 56 from 58, getting 2
No more digits to bring down!
6. If finding the last quotient digit resulted
in a non-zero remainder, write it next to
the quotient with a small r .
2 is the remainder; write r2 at the end of the quotient
“Round and Simplify” is used to get an estimated Quotient digit
(The estimated digit may need to be adjusted)
When divisors are longer than 1 digit, MrV’s Table can’t be used without doing some work area computations:
The “Round and Simplify” technique is used to get an estimated digit by using MrV’s Table.
“Check for Fit” uses multiplications to see if the estimated digit should be used , or be adjusted. For example:
64
34 2194
 204
154
136
18
r18
34 rounds down to 30, and rounding 219 to the 10’s digit rounds up to 220. (show work!)
Shorten to Simplify: Remove the matching right-hand zeroes from both to make a 1 digit divisor.
Row 3 with product 22 yields 7 as the estimated digit
 30 220
34 154
 3 22  7
 30 150
2
34
7
34
6
2 38
2 04
219  238 219  204
too big
 15 ok
 3 15  5
2
34
5
170
1
34
4
136
154  170 154  136
too big
34 does not go into 2 or 21, but it will divide into 219 – so the 1st quotient digit goes above the 9.
Let’s call 219 a partial dividend, the number that will have a product subtracted from it.
2. Round the divisor so that the leading digit is followed by all 0’s, and
round the partial dividend to the same position… Then shorten both to simplify.
Now you can use MrV’s Table to find the estimated quotient digit.
34 219
2
1. Find the place for the first quotient digit:
 18 ok
3. “Check the Estimated Digit for Fit: Is it too big, too small, or just right(Fit)?
- Multiply the estimated digit times the real divisor to make a test product:
a.) If the test product is > the partial dividend,
the estimated digit is too big: Subtract 1 from the estimated digit, and try again.
- Otherwise, subtract the test product from the partial dividend:
b.) If that number is < or = the real divisor, the estimated digit is just right!
c.) But if that number is > the real divisor,
the digit is too small: Subtract 1 from the estimated digit and try again.
- 7 times 34 makes 238 as a test product.
a.) Since 238 > 219, we need to reduce 7 to 6 and try again
b.) 6 times 34 makes 204 as a test product.
Since 204 < 219 and 219 – 204 = 15, 6 is just right: Use it as the exact quotient digit.
Bringing down the 4 makes 154 the new partial dividend.
4. Now let’s finish off this division by finding the 2nd quotient digit and the
remainder: Use Round and Simplify on 34 and 154…
When the divisor has two digits
(You will need a work area, and one row from MrV’s Times Table)
1. Find the position of
the first quotient digit:
18 does not go into 1 or 12, but it will
divide into 123 – so the 1st quotient
digit has to go right above the 3.
2. Estimate the value of
the first quotient digit:
Round both 18 and 123 to 10’s.
Shorten 20 and 120 to 2 and 12
Use MrV’s Table: Row 2 at 12 equals 6.
3. Check to see if the
estimated quotient digit fits:
Multiply 18 times 6, and get 108.
This is ok so far, since 108 is < 123.
6 is good, maybe 7 is better…
But 18•7 = 126, which exceeds 123,
so 6 is the best fit. Put 6 in the quotient
4. Move the quotient digit and
the product to the long division,
and subtract:
Put 108 under 123 and subtract it.
You get 15, which is good… it’s < 18.
6_ 8
????
18 1234
 108
154
 144
10
r10
Find 1st digit
Find 2 nd digit
12 3 down 120 12

18 up 20 2
15
15 4 down 150 15

18 up 20 2
4
18
6
108
5
18
7
126
6
18
8
144
5. Bring down the next digit from the
dividend, and repeat the process
Bring down 4 after the 15, making 154.
Round both 18 and 154 to 10’s.
Shorten 20 and 150 to 2 and 15.
In MrV’s Table, Row 2: 15 is between 14 and 16.
We already have 18x7=126, so do 18x8=144.
144 is closer to 154, so 8 is the right digit.
Put 8 above the 4 in 1234.
Put 144 under 154 and subtract it.
You get a remainder of 10; put it with the quotient.
6. For longer dividends,
repeat the process until the
last digit is processed. It may
come out evenly, or a
remainder may exist.
When a Quotient has an Internal or Ending Zero
(You need a work area, and the usual table row)
1. Find the position of
the first quotient digit:
31 does not go into 6, but it will
divide into 64 – so the 1st quotient
digit has to go right above the 4.
r12
?2_?,, 0? 9?0?
31 64,802
Round both 31 and 64 to 10’s.
Shorten 30 and 60 to 3 and 6
 62
MrV’s Row 3 at 6 gives quotient digit 2
28
3. Check to see if the
 00
estimated quotient digit fits:
Multiply 31 times 2, and get 62.
28 0
This is ok so far, since 62 is < 64.
 27 9
2 is the right digit because
64 – 62 = 2, which is smaller than 31
12
4. Bringing down the next digit may
 00
result in a partial dividend that is
still smaller than the divisor:
12
2. Estimate the value of
the first quotient digit:
Bring down 8 after the 2, making 28.
28 is still smaller than 31, so put a
zero as the 2nd quotient digit, put 00
under 28 and subtract getting 28.
Find 1st estimated digit
Find 1st estimated digit
31 64  30 60  3 6  2
31 280  30 280  3 28  9
31
31
2
9
62
279
5. Bring down the next digit from the
dividend, and repeat the process
Bring down 0 after the 28, making 280.
Round both 31 and 280 to 10’s.
Shorten 30 and 280 to 3 and 28.
Row 3: 28 is right after 27, which makes 9 the
3rd estimated quotient digit. 31 times 9 is 279,
so 8 is the right digit because 280 -279 = 1.
Bring down 2 after the 1, making 12.
12 is still smaller than 31, so put a zero as
the 4th (and last) quotient digit, put 00
under 12 and subtract getting 12.
12 is the remainder; Put r12 on the quotient.
Summary
• Using MrV’s Times Table for Long Division gives students an
excellent practice tool for understanding and performing the
process of doing Long Division, one digit at a time.
• If a student can do single digit multiplication and division in
his/her head, using MrV’s table may become unnecessary.
• Work areas are still needed when Round and Simplify needs to
be done for multiple-digit divisors.

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