New Math Mammoth books

There are now three new books available in the Blue Series. They are all for grades 5-6. The material for these books came from the Light Blue 5th and 6th grade curricula. Please click on the links to learn more and see samples.

1. 

   Math Mammoth The Four Operations
   (with a Touch of Algebra)

   The main topics studied in this book are simple equations, expressions that involve a variable, the order of operations, long multiplication, long division, and graphing simple linear functions.
   
   The idea is not to practice each of the four operations separately, but rather to see how they are used together in solving problems and in simple equations. We are trying to develop student's algebraic thinking. Many of the ideas in this chapter are preparing them for algebra in advance.
   
   

2. Math Mammoth Ratios & Proportions & Problem Solving

   This worktext concentrates, first of all, on two important concepts: ratios and proportions, and then on problem solving.
   
   First, we study thoroughly the concept of ratio, including how it connects with fractions and the aspect ratio in rectangles.
   
   Solving proportions is divided into three separate lessons. In the first one, we solve proportions by thinking through equivalent ratios. In the second one, the usual method of cross-multiplying is introduced. Then follows a lesson that explains just why cross-multiplying is allowed. Then there is more practice with solving proportions and word problems. We also study scaling geometric figures and floor plans.
   
   The last lessons concentrate on various kinds of word problems that can be solved with the help of bar or block diagrams (the same kind as in Singapore Math). These involve problems with fractional parts, and then problems involving ratios. The diagrams become a powerful tool to solve such problems without the use of algebra.
   
   
3. 

   Math Mammoth Place Value 5

   Math Mammoth Place Value 5 is a short worktext that covers value up to the trillions—that is, numbers up to 15 digits—rounding them, and using a calculator.
   
   The first lesson covers place value up to billions. Then, we study exponents, and right after that, place value up to trillions, writing the numbers in the expanded form using exponents. After working with addition a little, there are two lessons about calculator usage.
   
   

New Math Mammoth books

There are now three new books available in the Blue Series. They are all for grades 5-6. The material for these books came from the Light Blue 5th and 6th grade curricula. Please click on the links to learn more and see samples.

1. 

   Math Mammoth The Four Operations
   (with a Touch of Algebra)

   The main topics studied in this book are simple equations, expressions that involve a variable, the order of operations, long multiplication, long division, and graphing simple linear functions.
   
   The idea is not to practice each of the four operations separately, but rather to see how they are used together in solving problems and in simple equations. We are trying to develop student's algebraic thinking. Many of the ideas in this chapter are preparing them for algebra in advance.
   
   

2. Math Mammoth Ratios & Proportions & Problem Solving

   This worktext concentrates, first of all, on two important concepts: ratios and proportions, and then on problem solving.
   
   First, we study thoroughly the concept of ratio, including how it connects with fractions and the aspect ratio in rectangles.
   
   Solving proportions is divided into three separate lessons. In the first one, we solve proportions by thinking through equivalent ratios. In the second one, the usual method of cross-multiplying is introduced. Then follows a lesson that explains just why cross-multiplying is allowed. Then there is more practice with solving proportions and word problems. We also study scaling geometric figures and floor plans.
   
   The last lessons concentrate on various kinds of word problems that can be solved with the help of bar or block diagrams (the same kind as in Singapore Math). These involve problems with fractional parts, and then problems involving ratios. The diagrams become a powerful tool to solve such problems without the use of algebra.
   
   
3. 

   Math Mammoth Place Value 5

   Math Mammoth Place Value 5 is a short worktext that covers value up to the trillions—that is, numbers up to 15 digits—rounding them, and using a calculator.
   
   The first lesson covers place value up to billions. Then, we study exponents, and right after that, place value up to trillions, writing the numbers in the expanded form using exponents. After working with addition a little, there are two lessons about calculator usage.
   
   

Another algebra problem - or is algebra needed?

Updated!

There are some marbles in Box A and Box B. If 50 marbles from Box A and 25 from Box B are removed each time, there will be 600 marbles left in Box A when all marbles are removed from Box B. If 25 marbles from Box A and 50 marbles from Box B are removed each time, there will be 1800 marbles left in Box A when all marbles are removed from Box B. How many marbles are there in each box?

Again, this is from Singapore and teachers have told students not to use algebra to solve this question. However, any form of heuristic tools are allowed to facilitate the students in solving the questions.

I'd like to point out that I feel it's a good problem, but students might benefit from some "preparation". You could set up a preparation problem like this:

Jar A has 100 marbles and jar B has 40 marbles. You will start removing marbles one by one from jar A, but by 2's from jar B. How many marbles are left in jar A when jar B is empty?
What if you remove 2 marbles at a time from jar A and one marble at a time from B, then how many marbles are left in A when B is empty?

Then we can vary the numbers, for example let jar A have 250 and B have 90. Or, let A have 250 and B have 400 and see what happens! Lastly, change the number of marbles removed to 25 and 50 as in the real problem.

Now, the information given is "reversed" in the original problem because there we know how many marbles will be left and don't know how many marbles were there in the beginning. This does make the problem more difficult, obviously.


Solution:

When I saw this problem, my head automatically "saw" another usable unknown as "how many times do we scoop out marbles from each box?" You see, in scenario 1 we scoop out 50 marbles at a time from A and 25 from B, but the scooping is done the same number of times. So I called that n.

That automatically headed me down the "algebra" route... I wanted to write stuff using n:

There are 50n + 600 marbles in A, and 25n in B.

Then in situation 2, we don't do the same amount of scoopings... so the n is not the same. This time, we take 50 marbles at a time from B until it's empty, so B got emptied in double time as compared to situation 1. So... we have n/2 or half as many scoopings taking place.

Thus there are (n/2)*25 + 1800 in A, and (n/2)*50 in B.

Now you can get an equation that solves it by setting the number of marbles in A equal to number of marbles in A from the two situations:

50n + 600 = (n/2)*25 + 1800

Some solving... n = 32.

Therefore there are 50*32 + 600 = 2200 marbles in A, and 25 * 32 = 800 marbles in B.

---

Now, as far as solving it heuristically without using algebra... it was really hard for me at this point, since my head only wanted to consider the problem this one way. I started visualizing in my head two jars and two hands picking the marbles out... First one hand picks 50 each time out of the jar with more marbles and 25 out of the jar with less marbles until jar B runs out. Then the other way around: picking 25 out of A while taking 50 out of B, until B runs out.

Then I "saw" that the amount of marbles actually taken out from A in situation 1 was FOUR times the amount of marbles taken out from A in situation 2. (Just comparing how many marbles got taken out from A.)

This is because in 1, we took two times as many marbles out each time (50 vs 25), and also because in 1 it takes double that long (double the amount of scoopings) than in situation 2 (because we're timing all this by how quickly B runs out, and B runs out in half a time in situation 2).

We also know that the first time we took out 1200 more from jar A than in situation 2. So, that 1200 is 3/4 of the marbles taken out in sit 1. From which it's easy to get that 1200/3 * 4 = 1600 is the number of marbles taken out from A, in situation 1.

And that solves it then because now we know that there were 1600 + 600 = 2200 in A. And, since 1600 marbles were taken out from A by 50s, it means it was done 32 times. So B had 32 * 25 = 800 marbles.

---

I definitely think algebra is the easier way to go ... less brain strain for sure!

You can find yet other solution methods in the comments.

Another algebra problem - or is algebra needed?

Updated!

There are some marbles in Box A and Box B. If 50 marbles from Box A and 25 from Box B are removed each time, there will be 600 marbles left in Box A when all marbles are removed from Box B. If 25 marbles from Box A and 50 marbles from Box B are removed each time, there will be 1800 marbles left in Box A when all marbles are removed from Box B. How many marbles are there in each box?

Again, this is from Singapore and teachers have told students not to use algebra to solve this question. However, any form of heuristic tools are allowed to facilitate the students in solving the questions.

I'd like to point out that I feel it's a good problem, but students might benefit from some "preparation". You could set up a preparation problem like this:

Jar A has 100 marbles and jar B has 40 marbles. You will start removing marbles one by one from jar A, but by 2's from jar B. How many marbles are left in jar A when jar B is empty?
What if you remove 2 marbles at a time from jar A and one marble at a time from B, then how many marbles are left in A when B is empty?

Then we can vary the numbers, for example let jar A have 250 and B have 90. Or, let A have 250 and B have 400 and see what happens! Lastly, change the number of marbles removed to 25 and 50 as in the real problem.

Now, the information given is "reversed" in the original problem because there we know how many marbles will be left and don't know how many marbles were there in the beginning. This does make the problem more difficult, obviously.


Solution:

When I saw this problem, my head automatically "saw" another usable unknown as "how many times do we scoop out marbles from each box?" You see, in scenario 1 we scoop out 50 marbles at a time from A and 25 from B, but the scooping is done the same number of times. So I called that n.

That automatically headed me down the "algebra" route... I wanted to write stuff using n:

There are 50n + 600 marbles in A, and 25n in B.

Then in situation 2, we don't do the same amount of scoopings... so the n is not the same. This time, we take 50 marbles at a time from B until it's empty, so B got emptied in double time as compared to situation 1. So... we have n/2 or half as many scoopings taking place.

Thus there are (n/2)*25 + 1800 in A, and (n/2)*50 in B.

Now you can get an equation that solves it by setting the number of marbles in A equal to number of marbles in A from the two situations:

50n + 600 = (n/2)*25 + 1800

Some solving... n = 32.

Therefore there are 50*32 + 600 = 2200 marbles in A, and 25 * 32 = 800 marbles in B.

---

Now, as far as solving it heuristically without using algebra... it was really hard for me at this point, since my head only wanted to consider the problem this one way. I started visualizing in my head two jars and two hands picking the marbles out... First one hand picks 50 each time out of the jar with more marbles and 25 out of the jar with less marbles until jar B runs out. Then the other way around: picking 25 out of A while taking 50 out of B, until B runs out.

Then I "saw" that the amount of marbles actually taken out from A in situation 1 was FOUR times the amount of marbles taken out from A in situation 2. (Just comparing how many marbles got taken out from A.)

This is because in 1, we took two times as many marbles out each time (50 vs 25), and also because in 1 it takes double that long (double the amount of scoopings) than in situation 2 (because we're timing all this by how quickly B runs out, and B runs out in half a time in situation 2).

We also know that the first time we took out 1200 more from jar A than in situation 2. So, that 1200 is 3/4 of the marbles taken out in sit 1. From which it's easy to get that 1200/3 * 4 = 1600 is the number of marbles taken out from A, in situation 1.

And that solves it then because now we know that there were 1600 + 600 = 2200 in A. And, since 1600 marbles were taken out from A by 50s, it means it was done 32 times. So B had 32 * 25 = 800 marbles.

---

I definitely think algebra is the easier way to go ... less brain strain for sure!

You can find yet other solution methods in the comments.

Algebraic thinking

I downloaded the "balance" worksheet freebie, and daughter liked it. We homeschool and she would be in fifth grade this year if she were in public school. ... My question is about the balance worksheets - where would there be more of that? Stuff that does groundwork for algebraic thinking?

It's not just balance problems that prepare a child for algebra. These are important factors also:

• a good number sense (e.g. mental math)


• understanding of the four basic operations, for example how the opposite operations work. Another example: understanding that a division with remainder such as 50 ÷ 6 = 8 R2 is "turned around" with multiplication and addition: 8 × 6 + 2 = 50.


• a good command of fraction and decimal operations. Understanding the close connection between fractions and division.


• understanding the concepts of ratio and percent.

You can also simply write her more problems with an unknown. For example:

Write 7 + x = 28 and similar ones, like 12 + x = 99 and harder numbers.

x − 9 = 9 and notice how this is "solved" by adding.

6 − x = 4 and then harder numbers.

The same with multiplication and division.

I would also add one more thing that prepares children for algebra: good word problems -- not such that only require one operation to solve.

Singapore math's word problem booklets are told to be good, and here are some other (free) word problem websites.

In fact, soon I want to talk a little more about good word problems based on a paper I'm currently reading.

Algebraic thinking

I downloaded the "balance" worksheet freebie, and daughter liked it. We homeschool and she would be in fifth grade this year if she were in public school. ... My question is about the balance worksheets - where would there be more of that? Stuff that does groundwork for algebraic thinking?

It's not just balance problems that prepare a child for algebra. These are important factors also:

• a good number sense (e.g. mental math)


• understanding of the four basic operations, for example how the opposite operations work. Another example: understanding that a division with remainder such as 50 ÷ 6 = 8 R2 is "turned around" with multiplication and addition: 8 × 6 + 2 = 50.


• a good command of fraction and decimal operations. Understanding the close connection between fractions and division.


• understanding the concepts of ratio and percent.

You can also simply write her more problems with an unknown. For example:

Write 7 + x = 28 and similar ones, like 12 + x = 99 and harder numbers.

x − 9 = 9 and notice how this is "solved" by adding.

6 − x = 4 and then harder numbers.

The same with multiplication and division.

I would also add one more thing that prepares children for algebra: good word problems -- not such that only require one operation to solve.

Singapore math's word problem booklets are told to be good, and here are some other (free) word problem websites.

In fact, soon I want to talk a little more about good word problems based on a paper I'm currently reading.

Pan balance problems to teach algebraic reasoning

Today I have a "goodie" for you all: a free download of some pan balance (or scales) problems where children solve for the unknown:

Just right click on the link and "save" it to your own computer: Balance Problems (a PDF). This lesson is also included in my book Math Mammoth Multiplication 2 and in Math Mammoth Grade 4 Complete Worktext (A part).

The problems look kind of like this:

These can help children avoid the common misconception that equality or the equal sign "=' is an an operation. It is not; it is a relationship.

You see: many students view "=" as "find the answer operator", so that "3 + 4 = ?" means "Find what 3 + 4 is," and "3 + 4 = 7" means that when you add 3 and 4, you get 7. To students with this operator-view of equality, a sentence like "11 = 4 + 7" or "9 + 5 = 2 × 7;" makes no sense.

You might also find these resources useful:

Balance word problems from Math Kangaroo

Algebraic Reasoning Game - a weighing scales game that practices algebraic reasoning

Interactive Pan Balance with Shapes

A Balanced Equation Model from Absorb Mathematics
An interactive animation illustrating solving the equation 4x + 6 = x - 3. Drag the green handles to balance each side. Click the arrow button to reset the animation. On the right side, you'll see links to similar animations of equation solving using a balance.

Pan balance problems to teach algebraic reasoning

Today I have a "goodie" for you all: a free download of some pan balance (or scales) problems where children solve for the unknown:

Just right click on the link and "save" it to your own computer: Balance Problems (a PDF). This lesson is also included in my book Math Mammoth Multiplication 2 and in Math Mammoth Grade 4 Complete Worktext (A part).

The problems look kind of like this:

These can help children avoid the common misconception that equality or the equal sign "=' is an an operation. It is not; it is a relationship.

You see: many students view "=" as "find the answer operator", so that "3 + 4 = ?" means "Find what 3 + 4 is," and "3 + 4 = 7" means that when you add 3 and 4, you get 7. To students with this operator-view of equality, a sentence like "11 = 4 + 7" or "9 + 5 = 2 × 7;" makes no sense.

You might also find these resources useful:

Balance word problems from Math Kangaroo

Algebraic Reasoning Game - a weighing scales game that practices algebraic reasoning

Interactive Pan Balance with Shapes

A Balanced Equation Model from Absorb Mathematics
An interactive animation illustrating solving the equation 4x + 6 = x - 3. Drag the green handles to balance each side. Click the arrow button to reset the animation. On the right side, you'll see links to similar animations of equation solving using a balance.

Multiplying in parts and the standard algorithm

I haven't blogged for a while but I've been thinking about this topic for a little while now. It is your multiplication algorithm, also called long multiplication, or multiplying in columns. I also happen to be writing a lesson about it for my upcoming LightBlue series 4th grade book.

The standard multiplication algorithm is not awfully difficult to learn. Yet, some books advocate using so-called lattice multiplication instead. I assume it is because the standard method is perceived as being more difficult. But let's look at it in detail.

Before teaching the standard algorithm, consider explaining to the students multiplying in parts, a.k.a. partial products algorithm in detail:

To multiply 7 × 84, break 84 into 80 and 4 (its tens and ones). Then multiply those parts separately, and lastly add.

So we calculate the partial products first: 7 × 80 = 560 and 7 × 4 = 28. Then we add them: 560 + 28 = 588.

If you practice that for one whole lesson before embarking on the actual algorithm, how much better prepared the kids will be!

Next, they will see the standard way of multiplying:

2 
 84
×   7

588

Obviously, the steps here are the same. You multiply the ones first: 7 × 4 = 28, write down 8 of the ones, and carry the 2 of the tens. then you multiply 7 × 8 = 56, add 2 to get 58 and write that down in tens place.

What about this way of writing it down?

 84
×   7

28
+  560

588

It uses a little more space, but the underlying principle of multiplying in parts is more obvious.

It works with two two-digit numbers as well:

 84
×   47

28
560
160
3200

3948

Now, the individual multiplications are 7 × 4, then 7 × 80, then 40 × 4 and lastly 40 × 80.

Lastly, I'll touch on lattice multiplication. It uses the same exact principles; however I am not sure if it makes the underlying principle any more obvious to the students than the standard algorithm (and it does take more time and space).

    8   4  
  +---+---+
  |5 /|2 /|
  | / | / | 7
5 |/ 6|/ 8|
  +---+---+
    8   8

Answer 588.

Check out Lattice Multiplication to learn how it's actually done; it's hard to explain without images.

Either way, you NEED to explain multiplying in parts to the students. In this case it's not enough just to be able to go through the motions of an algorithm, because multiplying in parts is so needful in everyday life, and later in algebra (distributive property).

Consider for example these mental multiplications you might encounter while shopping:

5 × $14.
Just do 5 × $10 = $50 and 5 × $4 = $20, and add those. Answer $70. I'm sure most of us are quite used to doing such simple products mentally.

4 × $3.12. Go 4 × $3 = $12 and 4 × 12 ¢; = 48 ¢, and add. Answer $12.48.

Multiplying in parts and the standard algorithm

I haven't blogged for a while but I've been thinking about this topic for a little while now. It is your multiplication algorithm, also called long multiplication, or multiplying in columns. I also happen to be writing a lesson about it for my upcoming LightBlue series 4th grade book.

The standard multiplication algorithm is not awfully difficult to learn. Yet, some books advocate using so-called lattice multiplication instead. I assume it is because the standard method is perceived as being more difficult. But let's look at it in detail.

Before teaching the standard algorithm, consider explaining to the students multiplying in parts, a.k.a. partial products algorithm in detail:

To multiply 7 × 84, break 84 into 80 and 4 (its tens and ones). Then multiply those parts separately, and lastly add.

So we calculate the partial products first: 7 × 80 = 560 and 7 × 4 = 28. Then we add them: 560 + 28 = 588.

If you practice that for one whole lesson before embarking on the actual algorithm, how much better prepared the kids will be!

Next, they will see the standard way of multiplying:

2 
 84
×   7

588

Obviously, the steps here are the same. You multiply the ones first: 7 × 4 = 28, write down 8 of the ones, and carry the 2 of the tens. then you multiply 7 × 8 = 56, add 2 to get 58 and write that down in tens place.

What about this way of writing it down?

 84
×   7

28
+  560

588

It uses a little more space, but the underlying principle of multiplying in parts is more obvious.

It works with two two-digit numbers as well:

 84
×   47

28
560
160
3200

3948

Now, the individual multiplications are 7 × 4, then 7 × 80, then 40 × 4 and lastly 40 × 80.

Lastly, I'll touch on lattice multiplication. It uses the same exact principles; however I am not sure if it makes the underlying principle any more obvious to the students than the standard algorithm (and it does take more time and space).

    8   4  
  +---+---+
  |5 /|2 /|
  | / | / | 7
5 |/ 6|/ 8|
  +---+---+
    8   8

Answer 588.

Check out Lattice Multiplication to learn how it's actually done; it's hard to explain without images.

Either way, you NEED to explain multiplying in parts to the students. In this case it's not enough just to be able to go through the motions of an algorithm, because multiplying in parts is so needful in everyday life, and later in algebra (distributive property).

Consider for example these mental multiplications you might encounter while shopping:

5 × $14.
Just do 5 × $10 = $50 and 5 × $4 = $20, and add those. Answer $70. I'm sure most of us are quite used to doing such simple products mentally.

4 × $3.12. Go 4 × $3 = $12 and 4 × 12 ¢; = 48 ¢, and add. Answer $12.48.

Word problems in Singapore math and bar diagrams

If you're interested in Singapore math's word problems, and how their bar diagrams work, check out this blog entry by Denise at Let's Play Math:

Pre-algebra problem solving: 3rd grade

She goes through a bunch of word problems from Singapore Math 3-A book, and explains both a solution based on algebraic thinking, and a solution with bar diagrams.

It's good reading for all of us who teach, actually.

Word problems in Singapore math and bar diagrams

If you're interested in Singapore math's word problems, and how their bar diagrams work, check out this blog entry by Denise at Let's Play Math:

Pre-algebra problem solving: 3rd grade

She goes through a bunch of word problems from Singapore Math 3-A book, and explains both a solution based on algebraic thinking, and a solution with bar diagrams.

It's good reading for all of us who teach, actually.

Missing addend with x or a box?

Recently I've been writing worksheets for fourth grade... Later this year, I will make this collection available for whoever wishes to buy it. But for now, it's a work in progress.

One topic I was pondering a lot was whether to include problems such as

15 + x = 30 into the worksheets.
Or, 234 + x = 700
or x + 1,923 = 5,000.

I hope you get the idea; in elementary books you often see these kind of missing addend problems with a little box:

15 + = 30 or
234 + = 700 or
+ 1,923 = 5,000.

Well, I decided for the x over the box! I don't think solving missing addend problems with subtraction is too difficult to learn on fourth grade; after all, students have been working with addition and subtraction connection from 1st grade on, right?

In my own old schoolbooks I actually see x from 3rd grade on.

I made several problems with charts such as

Write a missing addend sentence
and a subtraction sentence to solve it.

         1,500

|------------|-----|
     x         346

Hopefully those help the students with algebraic thinking.

Tags: math, elementary, algebra

Missing addend with x or a box?

Recently I've been writing worksheets for fourth grade... Later this year, I will make this collection available for whoever wishes to buy it. But for now, it's a work in progress.

One topic I was pondering a lot was whether to include problems such as

15 + x = 30 into the worksheets.
Or, 234 + x = 700
or x + 1,923 = 5,000.

I hope you get the idea; in elementary books you often see these kind of missing addend problems with a little box:

15 + = 30 or
234 + = 700 or
+ 1,923 = 5,000.

Well, I decided for the x over the box! I don't think solving missing addend problems with subtraction is too difficult to learn on fourth grade; after all, students have been working with addition and subtraction connection from 1st grade on, right?

In my own old schoolbooks I actually see x from 3rd grade on.

I made several problems with charts such as

Write a missing addend sentence
and a subtraction sentence to solve it.

         1,500

|------------|-----|
     x         346

Hopefully those help the students with algebraic thinking.

Tags: math, elementary, algebra

Remainder in division

I was updating my Division 1 ebook and thought I'd share a lesson idea.

Have you ever tried this kind of exercise when studying remainder in division?

1 ÷ 3 = 0, R 1 2 ÷ 3 = 0, R 2 3 ÷ 3 = __, R __ 4 ÷ 3 = __, R __ 5 ÷ 3 = __, R __ 6 ÷ 3 = __, R __ 7 ÷ 3 = __, R __ 8 ÷ 3 = __, R __ 9 ÷ 3 = __, R __

10 ÷ 3 = __, R __ 11 ÷ 3 = __, R __ 12 ÷ 3 = __, R __ 13 ÷ 3 = __, R __ 14 ÷ 3 = __, R __ 15 ÷ 3 = __, R __ 16 ÷ 3 = __, R __ 17 ÷ 3 = __, R __ 18 ÷ 3 = __, R __

19 ÷ 3 = __, R __ 20 ÷ 3 = __, R __ 21 ÷ 3 = __, R __ 22 ÷ 3 = __, R __ 23 ÷ 3 = __, R __ 24 ÷ 3 = __, R __ 25 ÷ 3 = __, R __ 26 ÷ 3 = __, R __ 27 ÷ 3 = __, R __

You should do it with different divisors, and ask the student(s) to find a pattern. Do you know it?

Tags: math, elementary, lesson

Remainder in division

I was updating my Division 1 ebook and thought I'd share a lesson idea.

Have you ever tried this kind of exercise when studying remainder in division?

1 ÷ 3 = 0, R 1 2 ÷ 3 = 0, R 2 3 ÷ 3 = __, R __ 4 ÷ 3 = __, R __ 5 ÷ 3 = __, R __ 6 ÷ 3 = __, R __ 7 ÷ 3 = __, R __ 8 ÷ 3 = __, R __ 9 ÷ 3 = __, R __

10 ÷ 3 = __, R __ 11 ÷ 3 = __, R __ 12 ÷ 3 = __, R __ 13 ÷ 3 = __, R __ 14 ÷ 3 = __, R __ 15 ÷ 3 = __, R __ 16 ÷ 3 = __, R __ 17 ÷ 3 = __, R __ 18 ÷ 3 = __, R __

19 ÷ 3 = __, R __ 20 ÷ 3 = __, R __ 21 ÷ 3 = __, R __ 22 ÷ 3 = __, R __ 23 ÷ 3 = __, R __ 24 ÷ 3 = __, R __ 25 ÷ 3 = __, R __ 26 ÷ 3 = __, R __ 27 ÷ 3 = __, R __

You should do it with different divisors, and ask the student(s) to find a pattern. Do you know it?

Tags: math, elementary, lesson