Adding and subtracting within 20 from memory
There are a hundred sums of two one-digit numbers. How does a child come to know every one of them, and every subtraction that undoes one, without counting?
Where you are 7 in Adding and subtracting within 20
When you reach for it
Every adding and subtracting job from here on is built out of small facts. 46 + 38 needs 6 + 8 and 4 + 3; 52 - 27 needs 12 - 7. A child who has to count out 6 + 8 on fingers in the middle of a bigger problem often loses track of the bigger problem. So the aim now is that every sum of two one-digit numbers, and every subtraction that undoes one, is known from memory.
“From memory” has two honest forms. Some facts are simply recalled: asked 5 + 5, the child says 10, and nothing happens in between. Others arrive by a route so short it runs in the head in a moment: asked 9 + 6, the child thinks “nine needs one to make ten, five more, fifteen”. Both count. What does not count yet is counting by ones. It is correct and always allowed, but it is slow, and 8 + 7 by ones has seven steps to slip on. Routes that are used often turn into recall on their own.
The routes were each taught in a lesson of their own: counting on, doubles and near doubles, making ten, the bonds of ten, Fact families, and subtraction as a missing part. This lesson does not teach them again. It puts all the facts in one picture, matches each fact with a route that suits it, shows how one known fact reaches its neighbors, and turns every subtraction into an addition the child already knows.
The procedure
The lesson has four parts, one for each day: the whole set of facts, choosing a route, reaching a fact from one you know, and subtracting by thinking of adding. A short note at the end says how recall is built.
The whole set of facts
The facts to know are the sums of two one-digit numbers, from 0 + 0 to 9 + 9, and the subtractions that undo them, such as 17 - 8. Sums such as 10 + 7 or 13 + 4 are within 20 too, but they come from tens and ones (Ten more and ten less), not from memory work.
All the sums fit in one addition table. Choose a number down the left side and a number across the top. The box where that row and that column meet holds their sum. The box in row 7 and column 5 holds 12, because 7 + 5 = 12.
Make a large copy with the child and fill it in together, a row at a time, before reading anything off it. Then look for patterns. Each one is a fact about adding, not just about the table.
- Along a row the sums go up by one. Row 6 reads 6, 7, 8 and on to 15. One box to the right adds one more to the second number, so the sum is one more. This is why a fact’s neighbors are easy (the third part).
- The two sides of the shaded line match. The shaded boxes hold the doubles, 0 + 0 to 9 + 9. The box in row 3, column 8 holds 11, and so does the box in row 8, column 3, because 3 + 8 and 8 + 3 are the same amount: the order of the parts does not change the whole. This is the turnaround of Fact families, the commutative property of addition. It nearly halves the table.
- Each slanting line holds one sum. Run a finger from the top right toward the bottom left and the same number repeats. One line holds every 10, from 1 + 9 to 9 + 1: the bonds of ten. Along such a line one part goes up by one while the other goes down by one, so the whole stays the same.
- Row 0 and row 1 need no learning. Adding 0 changes nothing, and adding 1 gives the next counting number.
Children can color the table as facts become known. A table that fills with color over the weeks is a picture of what the child knows, with no score on it.
Choosing a route
Most facts have a route that suits them. Part of knowing a fact from memory is seeing at a glance which route fits, and running it without help.
| The fact looks like | A short route | For example |
|---|---|---|
| adding 0, 1, 2 or 3 | count on from the larger number | 8 + 3: nine, ten, eleven |
| both numbers the same | a double, known by heart | 7 + 7 = 14 |
| the numbers one apart | a double, then one more | 6 + 7 = 6 + 6 + 1 = 13 |
| a 9 or an 8, total over ten | make ten | 8 + 5 = 8 + 2 + 3 = 13 |
| total exactly ten | a bond of ten | 4 + 6 = 10 |
| ten and some more | tens and ones | 10 + 6 = 16 |
Each route has its own lesson, linked above; the job here is matching. Some facts fit two rows of the table. 8 + 9 is a near double and has a 9, so “8 + 8 and one more” and “make ten with the 9” both work, and both give 17. Any correct route is fine. The quickest one is the one with the fewest steps the child already knows, and that can differ from child to child.
Turnarounds are free. 3 + 8 is the same fact as 8 + 3, so a child who meets the small number first may swap the order and count on from the larger one.
Reaching a fact from one you know
A child does not need every fact by heart to answer every fact quickly. A few anchors, held firmly, reach the rest. Good anchors are the doubles, the bonds of ten, and ten plus a number.
One more, one less. If 6 + 6 = 12, then 6 + 7 is one more, because 7 is 6 and 1 more:
Grouping the parts this way is the associative property: the same numbers added in a different grouping make the same total. The step can be taken twice, so 6 + 8 is two more than 6 + 6, or taken down, so 6 + 5 is one less. In the table this is sliding along a row.
From ten. 10 + 5 = 15 is quick, because it is ten and five. 9 is one less than 10, so 9 + 5 is one less than 15, which is 14.
Giving one across. If one part gives 1 to the other, the whole does not change, since nothing was added or taken away. In the table this is one step along a slanting line. Making ten is this same move aimed at a 10; giving one across can aim at a double instead (Example 6).
Subtracting by thinking of adding
A subtraction within twenty asks for a missing part. 15 - 7 asks: 7 and what make 15? So the child thinks of the addition 7 + \square = 15. If 7 + 8 = 15 is known, the answer is 8. This is subtraction as a missing part, and here it becomes the main way subtraction facts are known at all. Counting back seven, or counting up by ones, gives the same answer more slowly.
So the addition facts carry the subtraction facts with them, and the addition table is a subtraction table too. To find 15 - 7, go along row 7 until you reach 15. The number at the top of that column, 8, is the answer. An addition fact with two different parts gives two subtraction facts: 7 + 8 = 15 gives 15 - 7 = 8 and 15 - 8 = 7, the fact family of the three numbers.
When the matching addition is not known yet, reach it from one that is, as in the third part (Example 8).
How recall grows
Recall grows from many short meetings with each fact, spread over many days, with the route said aloud at first. Some ways to give those meetings:
- Games with scores of their own. A beanbag toss, a card game where two cards are added, a board game moved by two number cubes. The adding is the point of the game.
- Talk about the route. Ask “how did you know?” as often as “what is it?”. A child who says “9 + 5 is 10 + 5, take one” is building the anchor and the route together.
- A few facts at a time. Work on the near doubles for a while, then the facts with a 9, coming back to earlier groups now and then.
- Sort the facts. The child sorts fact cards into “I just know it” and “I have a quick way”. The second pile shrinks on its own.
There is no clock in any of this. A fact is known when the child gives it correctly on different days and in different settings, and can say how they know. Speed then comes by itself, because a recalled fact takes no time to find.
Worked examples
Examples 1 and 2 go with the first part, 3 and 4 with the second, 5 and 6 with the third, and 7 and 8 with the fourth. Read each aloud with the child, with the table and some counters at hand.
Every way to make thirteen
Which boxes of the addition table hold 13? Write the fact for each.
SolutionNeither part can be 0, 1, 2 or 3: even 9 with 3 makes only 12. Start with the smallest part that works, 4, which needs 9. Then raise the first part by one and lower the second by one each time, which keeps the whole at 13:
The list stops at 9 + 4, because the next one would need a 10, and the table stops at 9. So 6 boxes hold 13, and they lie on one slanting line. They come in turnaround pairs, 4 + 9 with 9 + 4 and so on, so there are only 3 different facts here to know.
How many facts are there to learn?
The table has 10 rows and 10 columns. How many different facts are left to learn once turnarounds, adding 0 and adding 1 are taken out?
Solution10 rows of 10 boxes make 100 boxes. The 10 doubles on the shaded line have no turnaround partner. The other 90 boxes come in turnaround pairs, which is 45 pairs. So there are 10 + 45 = 55 different facts.
Take out the facts with a 0: 0 + 0, 0 + 1, and so on to 0 + 9, which is 10 facts. That leaves 45. Take out the facts with a 1 that are left: 1 + 1 to 1 + 9, which is 9 facts. That leaves 36.
So the real work is 36 facts, and many of those are doubles, near doubles or facts with a 9 that have short routes. It is worth showing a child: the job is much smaller than a hundred.
A route for each fact
Choose a route for each fact and find the sum: 9 + 3, 6 + 5, 8 + 2, 5 + 3.
Solution9 + 3 has a 9 and a total over ten, so make ten. 9 needs 1, and 3 is 1 and 2, so 9 + 3 = 10 + 2 = 12.
6 + 5 has its numbers one apart, so use a double: 5 + 5 = 10, and one more is 11. Or 6 + 6 = 12, and one less is 11.
8 + 2 is a bond of ten: 8 + 2 = 10.
5 + 3 adds a small number, so count on from 5: six, seven, eight. 5 + 3 = 8.
The point is not that each fact has one correct route. It is that a glance at the two numbers suggests a route of one or two steps.
Two beanbags
At the school fair, the beanbag board has five holes worth 5, 6, 7, 8 and 9 points. Each player tosses two beanbags, and both may land in the same hole. Tsegaye’s beanbags land in the 6 and the 9. What is his score?
The score is 6 + 9. Tsegaye turns it around to 9 + 6 and makes ten: 9 needs 1, and 6 is 1 and 5.
Abebech, watching, found it another way: “10 + 6 is 16, and 9 is one less than 10, so it is 15.” Both routes take two short steps. Counting on six from 9 also gives 15, but in six steps. Tsegaye scores 15 points.
Sliding along a row
Hinemoa knows 6 + 6 = 12 by heart. Use it to find 6 + 7, 6 + 8 and 6 + 4.
Solution7 is one more than 6, so 6 + 7 is one more than 12, which is 13. The picture shows why: the second row is the first row with one more light counter.
8 is two more than 6, so 6 + 8 is two more than 12, which is 14. 4 is two less than 6, so 6 + 4 is two less than 12, which is 10. That one is also a bond of ten, which checks it.
In the table all four facts sit in row 6, and each step to the right adds one to the sum.
Giving one across
Find 7 + 9 by turning it into a double.
SolutionTake 1 from the 9 and give it to the 7. The 7 becomes 8 and the 9 becomes 8:
The first step splits 9 into 1 and 8. The second adds the 1 to the 7 first, a regrouping the associative property allows. In the picture no counter is added or taken away; one just moves across the gap, so the whole is the same.
Check by making ten: 9 needs 1, and 7 is 1 and 6, so 7 + 9 = 10 + 6 = 16. Both routes agree.
Room on the shelf
A library shelf has room for 15 books. 7 books are on it now. How many more books will fit?
SolutionThis asks for a missing part: 7 books and how many more make 15? As an addition with a box,
7 + 8 = 15 is a near double, 7 + 7 = 14 and one more, so \square = 8. The subtraction is 15 - 7 = 8.
In the table, go along row 7 to the 15; it sits in column 8. So 8 more books will fit, and 15 - 8 = 7 comes free with it.
When the matching fact is not known yet
Find 17 - 8.
SolutionThink of adding: 8 + \square = 17. Suppose 8 + 9 is not known yet. A nearby double is: 8 + 8 = 16. That is one short of 17, so one more is needed, and 8 + 9 = 17:
This joins the last two parts: a known double reaches the addition fact, and the addition fact gives the subtraction. Check by adding back: 9 + 8 = 17.
Boundaries
- Knowing facts from memory covers sums of two one-digit numbers and the subtractions that undo them. 13 + 4 and 19 - 6 are within twenty, but they are found by tens and ones (the 3 and 4 make 7, so 17), not as separate facts.
- Subtraction here takes a part from a whole, so the first number must be at least as large as the second. 15 - 7 has an answer; 7 - 15 has no answer among the whole numbers 0, 1, 2, 3 and so on.
- The turnaround works for adding only. 8 + 5 = 5 + 8, but 8 - 5 = 3 while 5 - 8 has no whole-number answer.
- Giving one across keeps a sum, not a difference. 7 + 9 = 8 + 8, but 9 - 7 = 2 while 10 - 6 = 4. In a subtraction the difference stays the same when both numbers go up by one: 9 - 7 = 10 - 8 = 2.
- A route needs the facts it rests on. Making ten for 8 + 5 needs the bond 8 + 2 = 10 and the split of 5 into 2 and 3. A child without those counts on while the bonds are built.
Where this goes next
These facts are the parts every larger sum is made of. Adding two-digit numbers by tens and ones uses one fact in each place, and when the ones make more than ten, as in Regrouping in addition, the fact says how many ones are left over. Adding a one-digit number to a two-digit number by making a ten stretches making ten to sums such as 38 + 5, and each jump on an open number line is a known fact.
The shaded line leads straight to even and odd numbers. Every double is even, because it splits into two equal parts, and Counting by twos, fives, threes and fours runs through the same numbers 2, 4, 6, 8 and on.
Exercises
The whole set of facts
Use the addition table in the first part.
- Write every fact in the table that makes 11. How many boxes hold 11?
- Which sums appear in only one box of the table?
- Go along row 4. What sum is in column 6? In column 9?
- How many boxes hold a sum greater than 10?
Choosing a route
Choose a route, say why it fits, and find the sum.
- 9 + 4
- 3 + 4
- 2 + 9
- 7 + 3
- Abebech’s beanbags land in the 9 and the 5 on the board in Example 4. What is her score?
- Hinemoa scores 12 with her two beanbags. Which holes could they have landed in? Give every answer.
From a fact you know
- You know 5 + 5 = 10. Find 5 + 8 and 5 + 4.
- You know 10 + 8 = 18. Find 9 + 8 and 10 + 9.
- 8 + 6 = 7 + \square
- 4 + 9 = 5 + \square
- You know 4 + 4 = 8. Find 4 + 6 and 3 + 5.
Subtracting by thinking of adding
- 15 - 6
- 12 - 9
- 14 - 9
- 12 - 5
- A library shelf holds 13 books. Hinemoa takes 5 of them home. How many books are left on the shelf?
- Both of Abebech’s beanbags land in the 5. Tsegaye’s land in the 8 and the 9. How many more points does Tsegaye score than Abebech?
- 18 - \square = 9
Explain
- Why do the boxes on the two sides of the shaded line in the addition table hold the same numbers?
- Tsegaye knows 4 + 7 = 11. Which two subtraction facts does he know with it, and why?
- Is it always true that when one number goes up by one and the other goes down by one, the sum stays the same? Is it true for a difference? Give an example of each.
Answers
- 8 boxes: 2 + 9, 3 + 8, 4 + 7, 5 + 6, 6 + 5, 7 + 4, 8 + 3, 9 + 2. Neither part can be 0 or 1, since 9 + 1 = 10 is too small, and the list stops at 9 + 2.
- 0 and 18. The only way to make 0 is 0 + 0, and the only way to make 18 is 9 + 9, since 9 is the largest number in the table. Every other sum can be made at least two ways.
- 10 and 13. Row 4, column 6 is 4 + 6 = 10, a bond of ten; column 9 is 4 + 9 = 13, three boxes further along.
- 36. The sums 11 to 18 fill 8, 7, 6, 5, 4, 3, 2 and 1 boxes, and 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36.
- 13. There is a 9, so make ten: 9 + 1 + 3 = 10 + 3 = 13.
- 7. The numbers are one apart: 3 + 3 = 6, and one more is 7. Counting on 3 from 4 works too.
- 11. Turn it around and count on two from 9: ten, eleven.
- 10. A bond of ten: 7 and 3 make 10.
- 14 points. Make ten: 9 + 1 + 4 = 10 + 4 = 14.
- The 5 and the 7, or both in the 6. Hole 5 needs 7, hole 6 needs 6, hole 7 needs 5 (the same pair again), and holes 8 and 9 would need a 4 or a 3, which the board does not have.
- 13 and 9. 8 is three more than 5, so 5 + 8 is 10 + 3 = 13. 4 is one less than 5, so 5 + 4 is 10 - 1 = 9.
- 17 and 19. 9 is one less than 10, so 9 + 8 = 18 - 1 = 17. 9 is one more than 8, so 10 + 9 = 18 + 1 = 19.
- 7. 8 + 6 = 14, and giving one from the 8 to the 6 makes 7 + 7, which is still 14.
- 8. 4 + 9 = 13, and giving one from the 9 to the 4 makes 5 + 8, which is still 13.
- 10 and 8. 6 is two more than 4, so 4 + 6 = 8 + 2 = 10. 3 + 5 is 4 + 4 with one given across, so it is 8.
- 9. Think 6 + \square = 15; 6 + 9 = 15 by making ten with the 9.
- 3. Think 9 + \square = 12; 9 + 3 = 12.
- 5. Think 9 + \square = 14; 9 + 5 = 14.
- 7. Think 5 + \square = 12; 5 + 7 = 12, since 5 + 5 = 10 and two more.
- 8 books. 13 - 5: think 5 + \square = 13, and 5 + 8 = 13.
- 7 more points. Abebech scores 5 + 5 = 10 and Tsegaye 8 + 9 = 17. Think 10 + \square = 17: 10 + 7 = 17, so 17 - 10 = 7.
- 9. 18 - \square = 9 means 9 + \square = 18, and 9 + 9 = 18 is a double.
- The box in row a, column b holds a + b, and the box across the shaded line, in row b, column a, holds b + a. Both are the same two parts put together, only in a different order, so they make the same whole: the turnaround, or commutative property.
- 11 - 7 = 4 and 11 - 4 = 7. Both ask for a missing part of 11. If 4 and 7 make 11, then taking either part away leaves the other.
- For a sum, yes: one is moved from one part to the other, and nothing is added or taken away, as in 6 + 8 = 7 + 7 = 14. For a difference, no: 8 - 5 = 3 but 9 - 4 = 5.