Multiplication and Division Strategies
The goal of various strategies is to select a method for a particular set of numbers that is accurate, efficient, and
builds on number sense.
Traditional (You know this one)
35
x 28
Doubling and Halving (one of the numbers is doubled and the other is halved to make a problem that is easier to
solve)
35 x 28
70 x 14
70 x 10 = 700
70 x 4 = 280
700 + 280 = 980
Break Apart (students break the numbers apart in different ways)
35 x 28
35 x 20 = 700
35 x 5 = 175
35 x 3 = 105
700 + 175 + 105 = 980
or
10 x 28 = 280
10 x 28 = 280
10 x 28 = 280
5 x 28 = 140
280 + 280 + 280 + 140 = 980
or
30 x 20 = 600
5 x 20 = 100
30 x 8 = 240
5 x 8 = 40
600 + 100 + 240 + 40 = 980
Partial Product
35 x 28
30 x 20
30 x 8
20 x 5
5x8
600 + 240 + 100 + 40 = 980
or
35
x 28
600 (30 x 20)
100 (20 x 5)
240 (8 x 30)
40__(8 x 5)
980
Changing one number to make an easier problem
For the problem 35 x 28 first do 35 x 30 (this is really 2 groups of 35 too many)
35 x 30 = 1,050 - now compensate for the change by subtracting the two groups of 35
1,050 – 70 = 980
Area (Arrays)- Numbers can be broken up many ways. This strategy develops multiplication by multiples of 10 and
100 (if I know 3x8 then I know 30x8), the distributive property (used extensively in algebra), supports geometry
concepts of area, and can show the relationship between multiplication and division.)
This example shows an array with numbers broken apart by place value. (35 x 28)
30
5
8
30x20=600
30x8=240
20
5x20=100
40
30 x 20 = 600
30 x 8 = 240
5 x 20 = 100
5 x 8 = 40
Total
980
Cluster Problems-(helps show relationship between multiplication and division. Uses multiples of 10 and 100. Uses the idea “I
start with what I know to get to what I don’t know.”)
35 x 28
35 x 10 = 350
35 x 20 = 700
35 x 10 =350 – 2 groups of 35 = 280
700 + 280 = 980
OR
35 x 10 = 350
35 x 10 = 350
35 x 5 = 175 (this is half of 350)
35 x 1 = 35
35 x 2 = 70
Total
980
350 ÷ 10 = 35
700 ÷ 20 = 35
If I add 350 and 700 = 1,050 which is 30 groups of 35. I can take off 2 groups of 35 for a total of 70 off of 1,050. Now I have 980.
Lattice 35 x 28 (This strategy is no longer directly taught to students but some students use it to help them keep track of
the parts and where the numbers go.)
2
8
0
0
2
6
1
4
4
0
9
8
3
0
0
5
Add each of the
diagonals to get the
final product. The
diagonals are the
“places” in the
standard numbers.
28 x 35 = 980
980
Partial Quotients – Uses estimation, multiplication x10 and x100, students make fewer mistakes
The Partial Quotients Algorithm uses a series of
“at least, but less than” estimates of how many b’s are in a.
Students often begin with multiples of 10 because they’re
easiest.
There are at least ten 12’s in
158 (10 x 12=120). Record 10 as
the first estimate.
12
158
- 120
38
Subtract - 36
Subtract
There are at least three more
(3 x 12 = 36). Record 3 as the
next estimate.
2
10 (10 x 12=120)
3 (3 x 12=36)
13
(groups of 12)
Since 2 is less than 12, you can stop
estimating. The final result is the sum
of the guesses (10 + 3 = 13) plus what
is left over (remainder of 2 )
Here’s another one:
36
7,891
- 3,600
4,291
Subtract - 3,600
691
- 360
331
- 324
Subtract
7
100 (100 x 36=3600)
100 (100 x 36=3600)
10 (10 x 36=360)
9 (9 x 36=324)
219 R7