Math Mammoth Grade 5 Complete Curriculum now available

163 pages
147 lesson pages


218 pages
179 lesson pages

5-A contents and samples
5-B contents and samples

Finally it is ready! I know some folks have been waiting for the 5-B part to get finished, and now it is here!

Math Mammoth Grade 5 complete curriculum consists of two student worktexts (A and B), a separate answer key for each, chapter tests and an end-of-year test, cumulative reviews, and an easy worksheet maker (Internet access required) to make extra practice worksheets when needed.

The two books (worktext part A and part B) for 5th grade deal with

• multi-digit multiplication and long division
• simple equations
• problem solving
• place value with large numbers and the judicious use of calculator
• all operations with decimals
• statistics and graphing
• all fraction operations
• geometry: classifying and drawing triangles & quadrilaterals; calculating the area of rectangles, triangles, parallelograms, and compound figures; surface area and volume of rectangular prisms
• introduction to integers
• introduction to percent

You can purchase the whole curriculum as a download, or as printed copies.

The downloadable curriculum is available at Kagi and at Currclick. The printed copies are made by Lulu (see links on this page, under the cover image).

Math Mammoth Grade 5 Complete Curriculum now available

163 pages
147 lesson pages


218 pages
179 lesson pages

5-A contents and samples
5-B contents and samples

Finally it is ready! I know some folks have been waiting for the 5-B part to get finished, and now it is here!

Math Mammoth Grade 5 complete curriculum consists of two student worktexts (A and B), a separate answer key for each, chapter tests and an end-of-year test, cumulative reviews, and an easy worksheet maker (Internet access required) to make extra practice worksheets when needed.

The two books (worktext part A and part B) for 5th grade deal with

• multi-digit multiplication and long division
• simple equations
• problem solving
• place value with large numbers and the judicious use of calculator
• all operations with decimals
• statistics and graphing
• all fraction operations
• geometry: classifying and drawing triangles & quadrilaterals; calculating the area of rectangles, triangles, parallelograms, and compound figures; surface area and volume of rectangular prisms
• introduction to integers
• introduction to percent

You can purchase the whole curriculum as a download, or as printed copies.

The downloadable curriculum is available at Kagi and at Currclick. The printed copies are made by Lulu (see links on this page, under the cover image).

Dividing decimals

I feel students need to get grounded conceptually in this topic. So many times, all they learn about decimal division are the rules of how to go about decimal division when using long division, and it becomes an "empty" skill - a skill that lacks the conceptual foundation.

So for starters, we can do two different kinds of mental math division problems.

1. Division by a whole number - using mental math
   
   Here it is easy to think, "So much is divided between so many persons".
   
   0.9 ÷ 3 is like "You have nine tenths and you divide it between three people. How much does each one get?" The answer is quite easy; each person "gets" 0.3 or three tenths.
   
   And... remember ALWAYS that you can check division problems by multiplication. Since 3 × 0.3 = 0.9, we know the answer was right.
   
   0.4 ÷ 100 turns out to be an easy problem if you write 0.4 as 0.400:
   0.400 ÷ 100 is like "You have 400 thousandths and you divide it between 100 people; how much does each one get?" The answer is of course 4 thousandths, or 0.004. Check: 100 × 0.004 which is 100 × 4/1000 = 400/1000 or 0.400 = 0.4.
   
   Here are some more similar ones:
   
   0.27 ÷ 9
   
   0.505 ÷ 5
   
   0.99 ÷ 11
   ...and you can make more, just think of the multiplication tables.
   



2. Division where the quotient (answer) is a whole number
   
   This time it helps to think, "How many times does the divisor go into the dividend?" In these types of mental math problems, the answer ends up being a whole number. (Of course the teacher has to plan these problems just right.)
   
   For example, 0.4 ÷ 0.2. Ask, "How many times does 0.2 fit into 0.4?" The answer is, 2 times. So 0.4 ÷ 0.2 = 2. Again, we can check it by multiplying: 2 × 0.2 = 0.4.
   
   Other similar division problems to solve mentally:
   
   1 ÷ 0.5
   
   3 ÷ 0.5
   
   0.09 ÷ 0.03
   
   0.9 ÷ 0.1
   
   2 ÷ 0.4
   
   1 ÷ 0.01
   
   ...and so on.
   

This decimal division lesson taken from my Decimals 2 book illustrates these two kinds of mental division problems.

Towards the general case

After the student is familiar with the two special cases above, we can go forward and study decimal division problems in general. Even here, we will divide the problems into two classes, depending on whether the divisor is a whole number or not.

1. The divisor is a whole number.
   
   For example, 3.589 ÷ 4 or 0.1938 ÷ 83. These can simply be solved by long division as they are. Just put the decimal point in the same place in the quotient as where it is in the dividend.
   
   The "stumbling block" may come when the division is not even (this also leads into the study of repeating decimals). Generally, you can continue the division indefinitely by tagging zeros to the dividend, such as making 3.589 to be 3.589000. Then when you've continued the division as long as you wish (or as long as the book tells you to do it), cut the decimal off at a desired accuracy and round it.
   
   Typical problem in a textbook would say, "Do 2.494 ÷ 3 and give your answer with 3 decimal digits." For this, you need to do the long division until the fourth decimal digit - so as to be able to round to 3 decimal digits. Since 2.494 does not have four decimal digits, you tag a zero to it to make it have so (2.4940).
   
   Fortunately, this process is not generally difficult. It's the second case that's more of a problem.
   
   



2. The divisor is not a whole number.
   
   Here, we do something quite special before dividing, and turn the problem into one where the divisor is a whole number. Then, the actual division is done like explained above.
   
   I say this is special, because this special thing that we do is based on a very important general principle of arithmetic:
   
   If you multiply both the dividend and the divisor by some same number, the quotient won't change.
   
   Let's see it in action with some easy numbers:
   
   1000 ÷ 200 = 5
   
   100 ÷ 20 = 5
   
   10 ÷ 2 = 5
   
   Each time both the dividend and the divisor change by a factor of ten, but the quotient does not change.
   
   We can also try it using a factor of 3 (or any other number):
   
   8 ÷ 2 = 4
   24 ÷ 6 = 4
   72 ÷ 18 = 4
   
   Let's try one more time, with a factor of 2:
   
   30 ÷ 6 = 5
   
   15 ÷ 3 = 5
   
   7.5 ÷ 1.5 = 5
   
   3.75 ÷ 0.75 = 5
   
   H hopefully by now you have convinced the student(s) of this principle. Now we can apply it to those pesky decimal division problems.
   
   
   
   
   This image shows how the decimal division problem 0.644 ÷ 0.023 can be changed into another problem, with a whole number divisor, and with the same answer.
   
   In each step, we multiply both the dividend and the divisor by 10. This, of course, is the same process as moving the decimal point.
   
   Many textbooks only show the student the "trick" of moving the decimal point... but don't show him what that idea is based on.
   
   An example
   
   To solve 13.29 ÷ 5.19, we need to first change the problem so that the divisor 5.19 is a whole number. We multiply both the dividend and the divisor by 10 as many times as needful to accomplish that:
   
   13.29 ÷ 5.19
   = 132.9 ÷ 51.9
   = 1329 ÷ 519, and now off you go to do long division... I'm not saying it's the easiest long division problem in the world, since the divisor is 519. Let's try an easier one.
   
   
   2,916 ÷ 0.02
   = 29,160 ÷ 0.2
   = 291,600 ÷ 2 and now you can do the long division.
   
   Of course, in reality you can also multiply by 100 instead of taking two steps of multiplying by 10. But students can start out by multiplying by 10 as many times as needed.
   

Please also see the lesson on dividing decimals by decimals, from my Math Mammoth Decimals 2 book.

Dividing decimals

I feel students need to get grounded conceptually in this topic. So many times, all they learn about decimal division are the rules of how to go about decimal division when using long division, and it becomes an "empty" skill - a skill that lacks the conceptual foundation.

So for starters, we can do two different kinds of mental math division problems.

1. Division by a whole number - using mental math
   
   Here it is easy to think, "So much is divided between so many persons".
   
   0.9 ÷ 3 is like "You have nine tenths and you divide it between three people. How much does each one get?" The answer is quite easy; each person "gets" 0.3 or three tenths.
   
   And... remember ALWAYS that you can check division problems by multiplication. Since 3 × 0.3 = 0.9, we know the answer was right.
   
   0.4 ÷ 100 turns out to be an easy problem if you write 0.4 as 0.400:
   0.400 ÷ 100 is like "You have 400 thousandths and you divide it between 100 people; how much does each one get?" The answer is of course 4 thousandths, or 0.004. Check: 100 × 0.004 which is 100 × 4/1000 = 400/1000 or 0.400 = 0.4.
   
   Here are some more similar ones:
   
   0.27 ÷ 9
   
   0.505 ÷ 5
   
   0.99 ÷ 11
   ...and you can make more, just think of the multiplication tables.
   



2. Division where the quotient (answer) is a whole number
   
   This time it helps to think, "How many times does the divisor go into the dividend?" In these types of mental math problems, the answer ends up being a whole number. (Of course the teacher has to plan these problems just right.)
   
   For example, 0.4 ÷ 0.2. Ask, "How many times does 0.2 fit into 0.4?" The answer is, 2 times. So 0.4 ÷ 0.2 = 2. Again, we can check it by multiplying: 2 × 0.2 = 0.4.
   
   Other similar division problems to solve mentally:
   
   1 ÷ 0.5
   
   3 ÷ 0.5
   
   0.09 ÷ 0.03
   
   0.9 ÷ 0.1
   
   2 ÷ 0.4
   
   1 ÷ 0.01
   
   ...and so on.
   

This decimal division lesson taken from my Decimals 2 book illustrates these two kinds of mental division problems.

Towards the general case

After the student is familiar with the two special cases above, we can go forward and study decimal division problems in general. Even here, we will divide the problems into two classes, depending on whether the divisor is a whole number or not.

1. The divisor is a whole number.
   
   For example, 3.589 ÷ 4 or 0.1938 ÷ 83. These can simply be solved by long division as they are. Just put the decimal point in the same place in the quotient as where it is in the dividend.
   
   The "stumbling block" may come when the division is not even (this also leads into the study of repeating decimals). Generally, you can continue the division indefinitely by tagging zeros to the dividend, such as making 3.589 to be 3.589000. Then when you've continued the division as long as you wish (or as long as the book tells you to do it), cut the decimal off at a desired accuracy and round it.
   
   Typical problem in a textbook would say, "Do 2.494 ÷ 3 and give your answer with 3 decimal digits." For this, you need to do the long division until the fourth decimal digit - so as to be able to round to 3 decimal digits. Since 2.494 does not have four decimal digits, you tag a zero to it to make it have so (2.4940).
   
   Fortunately, this process is not generally difficult. It's the second case that's more of a problem.
   
   



2. The divisor is not a whole number.
   
   Here, we do something quite special before dividing, and turn the problem into one where the divisor is a whole number. Then, the actual division is done like explained above.
   
   I say this is special, because this special thing that we do is based on a very important general principle of arithmetic:
   
   If you multiply both the dividend and the divisor by some same number, the quotient won't change.
   
   Let's see it in action with some easy numbers:
   
   1000 ÷ 200 = 5
   
   100 ÷ 20 = 5
   
   10 ÷ 2 = 5
   
   Each time both the dividend and the divisor change by a factor of ten, but the quotient does not change.
   
   We can also try it using a factor of 3 (or any other number):
   
   8 ÷ 2 = 4
   24 ÷ 6 = 4
   72 ÷ 18 = 4
   
   Let's try one more time, with a factor of 2:
   
   30 ÷ 6 = 5
   
   15 ÷ 3 = 5
   
   7.5 ÷ 1.5 = 5
   
   3.75 ÷ 0.75 = 5
   
   H hopefully by now you have convinced the student(s) of this principle. Now we can apply it to those pesky decimal division problems.
   
   
   
   
   This image shows how the decimal division problem 0.644 ÷ 0.023 can be changed into another problem, with a whole number divisor, and with the same answer.
   
   In each step, we multiply both the dividend and the divisor by 10. This, of course, is the same process as moving the decimal point.
   
   Many textbooks only show the student the "trick" of moving the decimal point... but don't show him what that idea is based on.
   
   An example
   
   To solve 13.29 ÷ 5.19, we need to first change the problem so that the divisor 5.19 is a whole number. We multiply both the dividend and the divisor by 10 as many times as needful to accomplish that:
   
   13.29 ÷ 5.19
   = 132.9 ÷ 51.9
   = 1329 ÷ 519, and now off you go to do long division... I'm not saying it's the easiest long division problem in the world, since the divisor is 519. Let's try an easier one.
   
   
   2,916 ÷ 0.02
   = 29,160 ÷ 0.2
   = 291,600 ÷ 2 and now you can do the long division.
   
   Of course, in reality you can also multiply by 100 instead of taking two steps of multiplying by 10. But students can start out by multiplying by 10 as many times as needed.
   

Please also see the lesson on dividing decimals by decimals, from my Math Mammoth Decimals 2 book.

Math Mammoth Grade 5-A Complete Worktext

Finally! Grade 5-A is available for the LightBlue Series (the complete curriculum series). The part A of 5th grade focuses on

• multi-digit multiplication and long division
• simple equations
• problem solving
• place value with large numbers and the judicious use of a calculator
• all operations with decimals
• statistics and graphing

Please see the table of contents and samples for a complete lesson list, and read more info here.

Math Mammoth Grade 5-A Complete Worktext

Finally! Grade 5-A is available for the LightBlue Series (the complete curriculum series). The part A of 5th grade focuses on

• multi-digit multiplication and long division
• simple equations
• problem solving
• place value with large numbers and the judicious use of a calculator
• all operations with decimals
• statistics and graphing

Please see the table of contents and samples for a complete lesson list, and read more info here.

Worksheet news

Some worksheet-related news from HomeschoolMath.net site:

• Grade 5 worksheets - ready-made worksheets, yet different (randomly generated) each time.
  
  
• Decimal worksheets generator just got better. Now you can let the number of decimals vary randomly in the problems. Also includes a bunch of ready-made worksheets that you generate just by clicking on links.
  
  
• Addition worksheet generator got better also. Now you can set the range individually for addends 3-6, and randomly switch all the addends (previously only addends 1 and 2). This page also has some links to click on to make worksheets readily, without actually bothering with the generator itself.
  

Worksheet news

Some worksheet-related news from HomeschoolMath.net site:

• Grade 5 worksheets - ready-made worksheets, yet different (randomly generated) each time.
  
  
• Decimal worksheets generator just got better. Now you can let the number of decimals vary randomly in the problems. Also includes a bunch of ready-made worksheets that you generate just by clicking on links.
  
  
• Addition worksheet generator got better also. Now you can set the range individually for addends 3-6, and randomly switch all the addends (previously only addends 1 and 2). This page also has some links to click on to make worksheets readily, without actually bothering with the generator itself.
  

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

A simple ratio problem

Problem: If a:b = 1:3 and b:c = 3:4, find a:c.

Two ratios are given, third is to be found. This is very very simple. The picture shows the two given ratios as blocks.

We can see that a is one block and c is four blocks, so the ratio a:c is 1:4.

You don't need an image for that, of course, since the original ratios are so easy. If a:b=1:3 and b:c=3:4, b being the same in both cases, we can write the ratio a:b:c as 1:3:4 right off.

But what if the numbers weren't so friendly? What if it said this way:

If a:b = 1:3 and b:c = 5:7, find a:c.

This is solvable in various ways. I'll use equivalent ratios, in other words change the given ratios to equivalent ratios until we find ones where the b's are the same.

In the first ratio, 1:3, b is 3. In the other ratio, 5:7, it is 5. We can make those to be 15 by changing the ratios to equivalent ratios - which is done in an identical manner as changing fractions to equivalent fractions.

1:3 = 5:15 and 5:7 = 15:21.

Now the ratio of a:b:c is 5:15:21, so the asked ratio a:c is 5:21.

Squares that aren't squares?

Updated with solutions!

Today I want to highlight a square problem I saw at MathNotations. I hope Dave Marain doesn't mind me showing this picture and problem on my blog... I have no problem acknowledging it's from his blog. I COULD just tell you all to "go read it at Dave's blog....

BUT I don't feel that's the best way, IF I want you to think about this. I can just guess that most of the folks would feel too lazy to click on the link and go read it there (would you?). So I want to show it here.


Figures not drawn to scale! And this is important!

Now here's the question:

Does the given information in each diagram guarantee that each is a square?

If you don't think so, your mission is to draw a quadrilateral with the given information but that clearly does NOT look like a square.

The IDEA is to make our students THINK LOGICALLY, or practice their deductive reasoning skills. A great little problem.

The answers:


Figure 1 is not necessarily a square. The upper left corner angle can be of any size. The upper side can be of any length. And so on. See here two examples.

Figure two is not necessarily a square either since the "top" side can be of any length. But it is a rectangle.

Figure 3 in the original problem IS always a square!

Now, I'll write another post where we'll extend this idea to some parallelograms.

Squares that aren't squares?

Updated with solutions!

Today I want to highlight a square problem I saw at MathNotations. I hope Dave Marain doesn't mind me showing this picture and problem on my blog... I have no problem acknowledging it's from his blog. I COULD just tell you all to "go read it at Dave's blog....

BUT I don't feel that's the best way, IF I want you to think about this. I can just guess that most of the folks would feel too lazy to click on the link and go read it there (would you?). So I want to show it here.


Figures not drawn to scale! And this is important!

Now here's the question:

Does the given information in each diagram guarantee that each is a square?

If you don't think so, your mission is to draw a quadrilateral with the given information but that clearly does NOT look like a square.

The IDEA is to make our students THINK LOGICALLY, or practice their deductive reasoning skills. A great little problem.

The answers:


Figure 1 is not necessarily a square. The upper left corner angle can be of any size. The upper side can be of any length. And so on. See here two examples.

Figure two is not necessarily a square either since the "top" side can be of any length. But it is a rectangle.

Figure 3 in the original problem IS always a square!

Now, I'll write another post where we'll extend this idea to some parallelograms.

Clock face problem

Divide the face of the clock into three parts with two lines so that the sum of the numbers in the three parts are equal.

Let's first understand the problem. We need to draw lines into the clock face. It doesn't say the lines need to end in the center or go through the center. The lines could seemingly go many various ways.

Again, to help a student who can't even get started, just tell them to draw some lines into the clock face. Here's one example:



Of course this is not a solution since it does not fulfill the condition that the sum of the numbers in the three parts needs to be equal. But it gets you started. You have some lines, so you can ask the student to add the numbers in the three parts and verify the sums are NOT equal.

How do we make the sums to be equal? Just random trying would take a long time.

The KEY to this problem is that since the sum in each of the three parts is equal, if you add those three sums, you will get the sum of all the numbers in the clock face. Thus, we can find out the partial sum by computing the total sum and dividing by 3.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78. 78 ÷ 3 = 26.

So the sum of the numbers in each part has to be 26. After this key point, solving this problem is fairly easy.

Now, we need to find numbers on the clock face that will add up to 26. Again, trial and error will probably produce the result fairly soon. But we can use some thinking too:

Let's consider first the largest number on the clock, 12. What other numbers should go with 12 to make 26?
We can add lots of its neigbors to it and try.

12 + 1 + 2 + 3 + 4 + 5 = 27 so just adding numbers from 1 on in order won't work.

That means we might need to couple 12 and 11.

12 + 11 = 23. Need three more. So... 12 + 11 + 1 + 2 will give 26. That's the one part that we can separate with one line.

Then continuing from the larger ones, 10 + 9 + 8 = 27. This is too much. Need another approach.

But let's draw lines across the clock face so that 10 can be joined with some smaller numbers. You will see soon that 10 + 9 + 3 + 4 = 26.



The last ones left are 5 + 6 + 7 + 8 = 26.

Now we can try hunt for another solution. Is there one?

Note that we can only draw two lines. You can't do this in very many different ways. You can draw two lines across the clock face, or perhaps draw one line across and another from the circumference to the first line, as in my first picture. In the one solution we've found, the two parts consisted of neighboring numbers and the one middle part did not. In either case, we need to use some neighboring numbers to make up the sums.

When you consider the numbers from 1 to 12 on the clock face, the only neighboring numbers that add up to 26 are the ones we used: 12 + 11 + 1 + 2 and 5 + 6 + 7 + 8. So we have to use at least one of those sets for the one part. If you try it out, you will quickly note no other solutions are possible than the one found above.

This may not be considered a totally rigorous argument by mathematical standards, but it is important for students to become convinced that there are no other solutions. Convincing others of the same is the first step towards mathematical PROOF.

Clock face problem

Divide the face of the clock into three parts with two lines so that the sum of the numbers in the three parts are equal.

Let's first understand the problem. We need to draw lines into the clock face. It doesn't say the lines need to end in the center or go through the center. The lines could seemingly go many various ways.

Again, to help a student who can't even get started, just tell them to draw some lines into the clock face. Here's one example:



Of course this is not a solution since it does not fulfill the condition that the sum of the numbers in the three parts needs to be equal. But it gets you started. You have some lines, so you can ask the student to add the numbers in the three parts and verify the sums are NOT equal.

How do we make the sums to be equal? Just random trying would take a long time.

The KEY to this problem is that since the sum in each of the three parts is equal, if you add those three sums, you will get the sum of all the numbers in the clock face. Thus, we can find out the partial sum by computing the total sum and dividing by 3.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78. 78 ÷ 3 = 26.

So the sum of the numbers in each part has to be 26. After this key point, solving this problem is fairly easy.

Now, we need to find numbers on the clock face that will add up to 26. Again, trial and error will probably produce the result fairly soon. But we can use some thinking too:

Let's consider first the largest number on the clock, 12. What other numbers should go with 12 to make 26?
We can add lots of its neigbors to it and try.

12 + 1 + 2 + 3 + 4 + 5 = 27 so just adding numbers from 1 on in order won't work.

That means we might need to couple 12 and 11.

12 + 11 = 23. Need three more. So... 12 + 11 + 1 + 2 will give 26. That's the one part that we can separate with one line.

Then continuing from the larger ones, 10 + 9 + 8 = 27. This is too much. Need another approach.

But let's draw lines across the clock face so that 10 can be joined with some smaller numbers. You will see soon that 10 + 9 + 3 + 4 = 26.



The last ones left are 5 + 6 + 7 + 8 = 26.

Now we can try hunt for another solution. Is there one?

Note that we can only draw two lines. You can't do this in very many different ways. You can draw two lines across the clock face, or perhaps draw one line across and another from the circumference to the first line, as in my first picture. In the one solution we've found, the two parts consisted of neighboring numbers and the one middle part did not. In either case, we need to use some neighboring numbers to make up the sums.

When you consider the numbers from 1 to 12 on the clock face, the only neighboring numbers that add up to 26 are the ones we used: 12 + 11 + 1 + 2 and 5 + 6 + 7 + 8. So we have to use at least one of those sets for the one part. If you try it out, you will quickly note no other solutions are possible than the one found above.

This may not be considered a totally rigorous argument by mathematical standards, but it is important for students to become convinced that there are no other solutions. Convincing others of the same is the first step towards mathematical PROOF.

Solving the plus sign problem

How many addition signs should be put between digits of the number 987654321 and where should we put them to get a total of 99?

This is a fifth grade problem taken from Word Problems for Kids by Canada's SchoolNet.

The first step, as always, is to understand the problem. The student needs to know what is an "addition sign" and a "digit". We're simply asked to put plus signs in between those numbers and add them up, and try to come up with 99.

Then, after we have a basic idea of what the problem is about, is the time to do something. You know, often the child may say, "I don't know how to start. I have no idea what to do!"

But in this case, as often happens, you'll get somewhere as soon as you'll do something. It's really simple: put some plus signs in there and just see what happens. Let's simply put the plus sign in between every digit:

9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45.

Great, we got something. We got 45 which is too small. Now the student should start thinking HOW to make the sum bigger?

Obviously, the only way to do that is to use ONE OR SOME two-digit numbers. We need to omit at least one of those plus signs!

So try something. For example:

98 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 136.
9 + 8 + 7 + 6 + 5 + 4 + 3 + 21 = 63.

Encourage the student to try a few other possibilities. He/she should notice that if you make the two-digit number using the large digits, the sum will be MORE than if you use the smaller digits for that.

(That is, of course, really simple.)

At this point the student might simply use "brute force" and write out all the possibilities with one two-digit number. That's alright; that will get him an answer!

However, there is also a quicker way, if we use some thinking. The target answer 99 is about half-way in between the two sums I have above (136 and 63), so I would try next to use the "middle digits" such as 4, 5, and 6 as the two-digit number:

9 + 8 + 7 + 6 + 54 + 3 + 2 + 1 = 90, which is too little.
9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99, which is the right answer!

Any other possibility with one two-digit number will either be less than 90 or more than 99, so that is the only solution using one two-digit number and seven plus signs.

But, good problem solvers will also consider checking if this is the ONLY solution. There is the possibility of using TWO two-digit numbers. Indeed, I quickly stumbled upon another solution that way:

9 + 8 + 7 + 6 + 5 + 43 + 21 = 99.

It's easy to note that this is the only possibility using six plus signs, since if you put the plus sign between some other digits, your sum will be MORE than this sum.

I hope this is helpful to some of you. I know the problem is quite easy. My intention is simply to point out how a typical problem solving process can go. Observing the "tricks of the trade" can help you to solve problems, and to teach others do the same.

Solving the plus sign problem

How many addition signs should be put between digits of the number 987654321 and where should we put them to get a total of 99?

This is a fifth grade problem taken from Word Problems for Kids by Canada's SchoolNet.

The first step, as always, is to understand the problem. The student needs to know what is an "addition sign" and a "digit". We're simply asked to put plus signs in between those numbers and add them up, and try to come up with 99.

Then, after we have a basic idea of what the problem is about, is the time to do something. You know, often the child may say, "I don't know how to start. I have no idea what to do!"

But in this case, as often happens, you'll get somewhere as soon as you'll do something. It's really simple: put some plus signs in there and just see what happens. Let's simply put the plus sign in between every digit:

9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45.

Great, we got something. We got 45 which is too small. Now the student should start thinking HOW to make the sum bigger?

Obviously, the only way to do that is to use ONE OR SOME two-digit numbers. We need to omit at least one of those plus signs!

So try something. For example:

98 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 136.
9 + 8 + 7 + 6 + 5 + 4 + 3 + 21 = 63.

Encourage the student to try a few other possibilities. He/she should notice that if you make the two-digit number using the large digits, the sum will be MORE than if you use the smaller digits for that.

(That is, of course, really simple.)

At this point the student might simply use "brute force" and write out all the possibilities with one two-digit number. That's alright; that will get him an answer!

However, there is also a quicker way, if we use some thinking. The target answer 99 is about half-way in between the two sums I have above (136 and 63), so I would try next to use the "middle digits" such as 4, 5, and 6 as the two-digit number:

9 + 8 + 7 + 6 + 54 + 3 + 2 + 1 = 90, which is too little.
9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99, which is the right answer!

Any other possibility with one two-digit number will either be less than 90 or more than 99, so that is the only solution using one two-digit number and seven plus signs.

But, good problem solvers will also consider checking if this is the ONLY solution. There is the possibility of using TWO two-digit numbers. Indeed, I quickly stumbled upon another solution that way:

9 + 8 + 7 + 6 + 5 + 43 + 21 = 99.

It's easy to note that this is the only possibility using six plus signs, since if you put the plus sign between some other digits, your sum will be MORE than this sum.

I hope this is helpful to some of you. I know the problem is quite easy. My intention is simply to point out how a typical problem solving process can go. Observing the "tricks of the trade" can help you to solve problems, and to teach others do the same.

Math Mammoth Grade 5 Worksheets ready

I hope you're not tired of hearing this.... but this is what I've been busy with lately.

Just as of today, I got Math Mammoth Grade 5 worksheets ready and available for purchasing.

Like the others, there are two separate books, A and B, plus answer keys.

Price for the whole package is $10. And that includes 123 quality math worksheets all total.

Click the link to see sample worksheets.

And, I've also set up a volume discount for any of my math books:

For order totals at least $34 - a 20% discount.
For order totals at least $50 - a 25% discount.
For order totals at least $70 - a 30% discount.
Use coupon code 8A2301338 when ordering to get these discounts.

Math Mammoth Grade 5 Worksheets ready

I hope you're not tired of hearing this.... but this is what I've been busy with lately.

Just as of today, I got Math Mammoth Grade 5 worksheets ready and available for purchasing.

Like the others, there are two separate books, A and B, plus answer keys.

Price for the whole package is $10. And that includes 123 quality math worksheets all total.

Click the link to see sample worksheets.

And, I've also set up a volume discount for any of my math books:

For order totals at least $34 - a 20% discount.
For order totals at least $50 - a 25% discount.
For order totals at least $70 - a 30% discount.
Use coupon code 8A2301338 when ordering to get these discounts.