hu
nd
re
ten ds
s
on
es
1
5 4 6
+ 3 2 3
8 6 9
Think of each
column as being a
set. You need to
add up all the
numbers in each
set.
First add the ones
set, then add the
tens set, then add
the hundreds set,
and so on.
3
1
5 6
9 5
5 1
1
7
+ 5
13
1 1
5
3
+ 1
1 0
2
8
2
5
7
2
5
9
7
3
4
8
9
1
Longer columns of numbers will
mean larger numbers to regroup, but
you can still use the same process.
6
1
5
+ 9
15
Repeat the process for each set. If
the sum of a set is more than 9,
regroup by moving the sum’s tens
place digit over to the next column.
© 2008 abcteach.com
1
7
5
3
When you regroup on the last set, the
sum’s tens place digit will create a
new place value.
6
+ 5
11
Add the ones set. If the sum of that
set is more than 9, regroup by
moving the sum’s tens place digit
over to the next column.
1 1
7 5 6
+5 9 5
5 1
1
0
+0
1
5
2
1
7 5 6
+5 9 5
1
4
Another way to regroup
when adding a very long
column of numbers is to
make a mark above the
next column every time the
number goes over 9 (that
is, every time there is a
digit in the tens place), and
start again.
1
1 1
1 1
2
3
9
7
8
+ 1
3 2
5
2
8
2
5
7
9
hu
nd
re
ten ds
s
on
es
1
5 4 6
3 2 3
2 3 3
-
2
2 1
8 3 4
- 6 9 8
Think of each
column as being
a set. Subtract all
the numbers in
each set.
Begin by
subtracting the
ones set, then
the tens set, then
the hundreds set,
and so on.
When the digit on
the bottom is
larger than the
digit on the top,
you have to
regroup or
“borrow” from the
next column.
Borrow 1 from the tens place and
write the difference above it. Cross
out the original number. Send the
borrowed 1 to the top of the ones set.
3
2 14
8 3 4
- 6 9 8
6
14
- 8
6
Remember that the 1 you borrowed
from the tens place is really 1 group
of ten. Add that 10 to the number at
the top of the ones set.
Now, subtract the ones set.
© 2008 abcteach.com
4
7 12
2 14
12
- 9
3
8 3 4
- 6 9 8
3 6
Check each set before you subtract.
Regroup whenever the bottom digit is
larger than the top digit.
5
7 12
2 14
8 3 4
- 6 9 8
1 3 6
7
- 6
1
834 > 698
You should not have to regroup on
the last set, because in the original
problem, the top number should be
larger than the bottom number.
6
5 15
6 10
6 5 9 7 0
- 5 7 2 3 8
0 8 7 3 2
Sometimes, you will need to
regroup for some sets, but not
others.
ONES column:
6 + 5 = 11
write the 2 below
regroup the 1 to the tens column;
1
0
+0
1
1
8
5
4
1
9 6
3 6
3 2
TENS column:
1 + 9 + 3 = 13
write the 3 below
regroup the 1 to the hundreds column
HUNDREDS column:
1 + 8 + 5 = 14
write the 4 below
regroup the 1 to the hundreds column
THOUSANDS column:
1+0+0=1
ONES column:
2 - 6 = can’t do that
borrow from the tens column
Make the (left) 4 a 3; make the (right) 2 a 12
12 - 6 = 6
TENS column:
3 8 = can’t do that
borrow from the hundreds column
Make the (left) 1 a 0; make the (right) 3 a 13
13 - 8 = 5
HUNDREDS column:
1 - 5 = can’t do that
borrow from the thousands column
Make the (left) 2 a 1; make the (right) 0 a 10
10 - 5 = 6
© 2008 abcteach.com
1 10
0 13
3 12
2 1 4 2
- 1 5 8 6
5 5 6
THOUSANDS column:
1-1=0
To multiply multi-digit numbers, multiply each place value separately to get a partial
product. Then, add the products together. A shortcut to doing this is to regroup digits in
the partial products, and add them as you go.
2 2
598
x
3
1794
SHORT FORM:
Regrouping is
used and place
value is kept
during the
process.
3 x 8 = 24
regroup the 2
3 x 9 = 27 + 2 = 29
regroup the 2
LONG FORM:
Here is the same
problem without using
regrouping, to show
the partial products
added together to
make the product.
Notice the place values
of each number.
3 x 8 = 24
3 x 90 = 270
598
x
3
24
270
+1500
1794
3 x 500 = 1500
3 x 5 = 15 + 2 = 17
24 + 270 +1500 = 1794
When multiplying two multi-digit numbers, using the short form is much easier.
3 3
1 1
476
x 52
1 9 5 2
+2380
24752
© 2008 abcteach.com
First, multiply by the bottom ones column:
2 x 6 = 12, regroup the 1
2 x 7 = 14 + 1 = 15, regroup the 1
2x4=8+1=9
so, 2 x 476 = 952
Next, multiply by the bottom tens column:
Remember to begin your answer in the tens column
5 x 6 = 30, regroup the 3
5 x 7 = 35 + 3 = 38, regroup the 3
5 x 4 = 20 + 3 = 23
so, 5 x 476 = 2380
Then, add the partial products:
2+0=2
5+0=5
9 + 8 = 17, regroup the 1
3+1=4
2+0=2
so, 476 x 52 = 24752