Pattern & Algebra B
Years 4 - 10

A Mathematics Centre Resource
You are encouraged to contribute to these notes. Comments from teachers or students, photos and student work samples will enrich the professional experience for all of us.

Menu

Introduction

Print Resources

Materials

  • Square tiles or cube blocks
  • Popsticks
    (Short or cut straws or plastic sticks are an alternative to popsticks)
  • Poly Plug
    (Could be an alternative for the first three investigations)
  • See Equipment List

Picture Puzzles have multiple levels of success. They do not have to be 'finished'. They can be revisited and continued.

Introduction

This menu develops algebra visually. Each Picture Puzzle begins with an experiment involving a pattern that can be touched, seen and described in words and pictures. Each student is encouraged to see the pattern in their own way first and then show they can apply it to the 100 case.

When confident with their own view, students are challenged to view the same experiment through someone else's eyes. If they can, they are asked to explain and apply the alternative. Two or three alternatives to their own view are presented in each investigation and these include both a substitution and a working backwards (equation solving) question. This not only confirms that seeing mathematics in different ways is valued, but provides opportunity to compare them and perhaps see why mathematicians have a preference for the simplest algebraic expression.

If teachers choose to, the various views in each experiment and their symbolic representations can be a natural doorway into equivalent algebraic expressions. The experiments also provide ordered pairs / tables of values which can be used to explore graphical algebra and/or spreadsheets. Communication skills - oral, written and symbolic - are a feature of every puzzle.

 

Summary

Four pathways of square tiles (or is it two paths?) extend equally from a central square to make a shape like a plus sign (+). Students are asked to find the total number of tiles for small arm lengths, then calculate the number of tiles for an arm length of 100. This leads into the general question If I tell you any arm length can you tell me the number of tiles needed to make the shape?.

A mathematician asks Can I check it another way? and later in this Picture Puzzle Mattias, Sue, Bruno and Christine reveal their ways of seeing the pattern. Students are then invited to explain and calculate in these ways, one of which might be the one they were already using.

Answers


Answers will vary because the students are making their own size 4 Arm Shapes. However, you could encourage them to record arm length and total tiles.

The process of making the shapes encourages the brain to predict the number of tiles needed for other cases. This slide is only asking for an oral response but...

...this next one expects that to be written. The symbols don't have to be used, but since they are there, why not? ... as long as learners attribute meaning to them.
  • For an arm length of 100 the number of tiles is 401, but ways of explaining will vary.
  • Different ways of seeing this experiment are the focus of the remainder of the puzzle.

 


This slide is the last of a sequence that portrays Mattias as seeing one arm rotate around the centre to make the shape.

Now Mattias sees the central tile pop in.

Encourage students to say, then write in words, then code the words as symbols. Something like:

  • Mattias sees one arm appearing four times and then one added into the centre.
  • T = 4 x A + 1. Students might also write this as a sum. If so, encourage the multiplication.
  • If A = 100, T = 4 x 100 + 1 = 401
This is the backwards question. It might be answered something like this:

  • To work out the length of each arm, Mattias would use one for the centre, then divide the remainder by four.
  • A = (T - 1) 4.
  • If T = 73, subtract 1 to get 72, then divide by 4 to get 18.
    Each arm is 18 tiles.

 


Sue sees it differently. These two slides are the last of a sequence portraying Sue as beginning with one in the centre then building the arms one tile at a time in sets of four.

Although Sue's way and Mattias's way give the same total, they demonstrate that 4 x A and A x 4 are not 'the same thing'. The symbolic expressions represent two completely different ways of building the 4 Arm Shape.

Encourage students to say, then write in words, then code the words as symbols. Something like:

  • Sue sees one in the centre first, then she puts on 4 around the centre, then 4 more, then 4 more and so on.
  • T = 1 + A x 4. Students might also write this as a sum. If so, encourage the multiplication.
  • If A = 16, T = 4 x 16 + 1 = 65
This is the backwards question. It might be answered something like this:
  • To work out the length of each arm, Sue would keep one to use in the centre. Then she would find out how many groups of 4 there are in the remainder. That would tell her the arm length.
  • A = (T - 1) 4.
    Same equation as Mattias, but a different interpretation of division. Mattias is 'shared between' or 'divided between'. Sue is 'how many'.
  • If T = 37, keep 1 for the centre. That gives 36. Then find out how many 4s in 36.
    Each arm is 9 tiles.

 


Bruno sees one long section first which is made from two arm lengths and the central tile.

Then he sees another one of these across it to give the total. But he has to remember he has counted the central tile twice.

Encourage students to say, then write in words, then code the words as symbols. Something like:

  • Bruno sees two lots of two arm lengths plus one, then he has to subtract one central tile.
  • T = 2 x (2A + 1) - 1.
  • If A = 20
    T = 2 x (2 x 20 + 1) -1
       = 2 x 41 - 1
       = 82 - 1
       = 81
This is the backwards question. It might be answered something like this:
  • To work out the length of each arm, Bruno has to add in a tile for the central one that is counted twice. Then he has to divide that number by two to know the number of tiles in 2A + 1. Then he has to take off 1 and divide by two again to find A.
  • A = {[(T + 1) 2] - 1} 2.
    Mmm, working backwards with Bruno's equation is not as straight forward as the two above.
  • If T = 73
    A = {[(73 + 1) 2] - 1} 2
       = [(74 2) - 1] 2
       = (37 - 1) 2
       = 36 2 = 18

Christine sees the 4 Arm Shape as the central paths in a square.
  • She can easily calculate the total number of tiles in her imagined square.
  • It's the square of (two arm lengths plus 1).
  • She can also easily calculate the number of tiles in each of the corner squares.
  • It's the square of (one arm length).
  • A subtraction gives the total of tiles in the 4 Arm Shape.

Encourage students to say, then write in words, then code the words as symbols. Something like:
  • T = (2 x A + 1)2 - 4 x A2
  • If A = 98
    T = (2 x 98 + 1)2 - 4 x 982
       = 1972 - (2 x 98)2
       = (197 + 196)(197 - 196)
       = 393
    (see Square Numbers in Pattern & Algebra A for difference between two squares)

This backwards question is very likely to force Christine to rethink her method. As natural as it may be - especially to anyone with recent experience in calculating area - it is not clear how it can be used to tackle the inverse problem. Christine might need to find an equivalent to her equation in order to calculate arm length from tiles. This may be an opportunity to suggest simplifying her equation:

T = (2 x A + 1)2 - 4 x A2
   = (2A + 1)2 - (2A)2
   = (2A + 1 + 2A)(2A + 1 - 2A)
   = 4A + 1 (rather neat really)

Which is simple for both substitution (as above for an arm length of 98) and for solution. If T = 77, the equivalent equation to Christine's rule shows the A = 19.

This example supports a case for valuing the simplest algebraic form of an expression.
(Note: If your staff or students develop a plausible backwards method using Geoffrey's equation, we will be happy to publish it here.)

To learn more about this Picture Puzzle click the photo to visit the Task Cameo.

Summary

A garden bed of plants in one row is surrounded by a path of paving tiles. Students are asked to find the total number of tiles for small numbers of plants, then calculate the number of tiles for a garden bed of 100 plants. This leads into the general question If I tell you any number of plants can you tell me the number of tiles needed to surround it?.

A mathematician asks Can I check it another way? and later in this Picture Puzzle Rashida, Steven, Irene and Geoffrey reveal their ways of seeing the pattern. Students are then invited to explain and calculate in these ways, one of which might be the one they were already using.

Answers


Answers will vary because the students are making their own garden bed lengths. However, you could encourage them to record plant and tile numbers.

The process of making the garden beds encourages the brain to predict the number of tiles needed for other cases. This slide is only asking for an oral response but...

...this next one expects that to be written. The symbols don't have to be used, but since they are there, why not? ... as long as learners attribute meaning to them.
  • For 100 plants the number of tiles is 206, but ways of explaining will vary.
  • Different ways of seeing this experiment are the focus of the remainder of the puzzle.

 


This slide is the last of a sequence that portrays Rashida as seeing the garden grow one plant and two tiles at a time.

Then she adds the six for the ends.

Encourage students to say, then write in words, then code the words as symbols. Something like:

  • Rashida builds the garden one plant at a time. Each plant needs two tiles. Then she adds six more for the ends.
  • T = P x 2 + 6.
  • If P = 81
    T = 81 x 2 + 6
       = 162 + 6
       = 168
This is the backwards question. It might be answered something like this:

  • To work out the number of plants, Rashida subtracts six for the ends, then works out how many twos can be made with what's left.
  • P = (T - 6) 2.
  • If T = 40, subtract 6 to get 34, then divide by 2 to get 17.
    She would need 17 plants to make the garden.

 

  • Steven sees that if the garden is only one plant, it needs eight tiles. Then it's like the rest of the plants make a long line with the same number of tiles above and below.
  • T = 8 + 2 x (P - 1).
  • If P = 102
    T = 8 + 2 x (102 - 1)
       = 8 + 2 x 101
       = 8 + 202
       = 206

This is the backwards question. It might be answered something like this:

  • To work out the number of plants, Steven subtracts eight needed for the ends and the first plant, then divides what's left by two.
  • P = (T - 8) 2.
  • If T = 30, subtract 8 to get 22, then divide by 2 to get 11.
    He would need 11 plants to make the garden.

 

  • Irene sees two L-shapes. Each one is as long as the plant line plus three more for the ends.
  • T = 2 x (P + 3).
  • If P = 54
    T = 2 x (54 + 3)
       = 2 x 57
       = 114

This is the backwards question. It might be answered something like this:

  • To work out the number of plants, Irene divides the number or tiles by two to get the tiles in an L-shape, then subtracts 3 for the short part of the L.
  • P = (T 2) - 3.
  • If T = 110
    P = (110 2) - 3
       = 55 - 3
       = 52.
    She needs 52 plants to make the garden.

 

  • Geoffrey imagines first imagines that all the squares are tiles. Then he can easily multiply the length of one row by three. One row is as long as the number of plants plus 2. After multiplying he takes way the number of plants and the tiles must be the number left.
  • T = 3 x (P + 2) - P.
  • If P = 67
    T = 3 x (67 + 2) - 67
       = 3 x 69 - 67
       = 207 - 67
       = 140

This backwards question is very likely to force Geoffrey to rethink his method. As natural as it may be - especially to anyone with recent experience in calculating area - it is not clear how it can be used to tackle the inverse problem. Geoffrey might need to find an equivalent to his equation in order to calculate plants from tiles. This may be an opportunity to suggest simplifying his equation:
T = 3 x (P + 2) - P
   = 3P + 6 - P
   = 2P + 6
Which is simple for both substitution (as above for 67 plants) and for solution. If T = 60, the equivalent equation to Geoffrey's rule shows the P = 27.

This example supports a case for valuing the simplest algebraic form of an expression.
(Note: If your staff or students develop a plausible backwards method using Geoffrey's equation, we will be happy to publish it here.)

To learn more about this Picture Puzzle click the photo to visit the Task Cameo.

Summary

You are told which position you have in a line. It is the same from both ends. You are asked to work out the number of people in the line. First students are asked to find the total number of people given small position numbers, then calculate the number of people if they are told they are 100th from each end. This leads into the general question If I tell you any position from each end, can you tell me the total number of people in the line?.

A mathematician asks Can I check it another way? and later in this Picture Puzzle Evo, Maya, Ishan and Ameli reveal their ways of seeing the pattern. Students are then invited to explain and calculate in these ways, one of which might be the one they were already using.

Answers


23 people in the line. Explanations will vary.

11 people in the line. Explanations will vary.

37 in the line. Explanations will vary.

If you are 100th from each end there are 199 people in the line. Again explanations will vary from student to student, however, the next part of the puzzle invites students to consider alternative explanations for working out the total.

 

  • Evo sees himself in two equal groups and his position counts the number of people in the group. So he can double his position number and then subtract one because he has counted himself twice.
  • N = 2 x P - 1, where N = number of people and P = position number.
  • If P = 35
    N = 2 x 35 - 1
       = 70 - 1
       = 69

  • Evo has to add one first (so there are two of 'him') then divide that number by two. That will tell him his position in the line.
  • P = (N + 1) 2
  • If N = 29
    P = (29 + 1) 2
       = 30 2
       = 15
    He is 15th in the line.

 

  • Maya sees herself as between to sets of people. Each set has the same number of people. That number is one less than her position number.
  • N = 2 x (P - 1) + 1, where N = number of people and P = position number.
  • If P = 50
    N = 2 x (50 - 1) + 1
       = 2 x 49 + 1
       = 98 + 1
       = 99

  • Maya has to take one away from the number (that's herself), then divide that number by two. Then adds one to that to count herself again to find her position in the line.
  • P = [(N - 1) 2] + 1
  • If N = 107
    P = [(107 - 1) 2] + 1
       = (106 2) + 1
       = 53 + 1
       = 54
    She is 54th in the line.

 

  • Ishan sees himself in one set of people, but not in the other. The set he is in is has one more person (himself) than the other set. His set is counted by the position number. The other set is not.
  • N = P + (P - 1), where N = number of people and P = position number.
  • If P = 84
    N = P + (P - 1)
       = 84 + (84 - 1)
       = 84 + 83
       = 167

  • Ishan has to find two numbers that are one apart and add to the total. He could guess and check or he could halve the total (which is always odd) and use the whole numbers on either side. His position number would be the higher of the two.
  • P = Int (N 2) + 1/2, where Int means integer value of...
  • If N = 85
    P = Int (N 2] + 1/2
       = Int (85 2) + 1/2
       = 421/2 + 1/2
       = 43
    He is 43rd in the line.

 

  • Ameli has seen herself as a pivot and imagined one set of people rotated 90 degrees around her. Then she is in both lines and her position number counts the length and width of a square. Cleverly, she now imagines the square completely filled with people. The imagined people make a square that is one less in length and width than 'her' square. If she calculates both squares and subtracts one from the other, the people left must be the number in the original line. Phew!
  • N = P2 - (P - 1)2, where N = number of people and P = position number.
  • If P = 90
    N = 902 - (90 - 1)2
       = 902 - 892
       = 8100 + 7921 ... or ... (90 + 89)(90 -89)
       = 179

    See Square Numbers in Pattern & Algebra A for difference between two squares.

  • To work backwards using this thinking, Ameli has a challenge. She needs to find two numbers that are one apart and when she squares them their difference is the total number of people. However, she does know that her position will be in the middle of the line, so she could start by halving the total (which will always result in a half number) and using the numbers on either side of this middle. Consider the example on the slide.
  • If N = 37
    • Half of this is 181/2.
    • Choose 19 and 18 and square them to get 361 and 324.
    • 361 - 324 = 37 as required. Great guess!
    • She must be 19th from either end.

To learn more about this Picture Puzzle click the photo to visit the Task Cameo.

Summary

Matches are arranged in a chain of triangles just like an engineering company might do before welding them together to make a support structure. First students are asked to find the total number of matches given small numbers of triangles, then calculate the number matches for 100 triangles. This leads into the general question If I tell you any number of triangles, can you tell me the total number of matches needed to make them?.

A mathematician asks Can I check it another way? and later in this Picture Puzzle Camilla, Keith, Mia and Waleed reveal their ways of seeing the pattern. Students are then invited to explain and calculate in these ways, one of which might be the one they were already using.

Answers


Answers will vary because the students are making their own chains. However, you could encourage them to record triangle and match numbers.

The process of making the triangle chains encourages the brain to predict the number of matches needed for other cases. This slide is only asking for an oral response but...

...this next one expects that to be written. The symbols don't have to be used, but since they are there, why not? ... as long as learners attribute meaning to them.
  • For 100 triangles the number of matches is 201, but ways of explaining will vary.
  • Different ways of seeing this experiment are the focus of the remainder of the puzzle.

 

Camilla sees the chain as beginning with one match. She completes the first triangle with two matches, then continues to make triangles by adding pairs of matches as far as required. Encourage students to say, then write in words, then code the words as symbols. Something like:
  • M = 1 + T x 2
  • If T = 41
    M = 1 + 41 x 2
         = 1 + 82
         = 83

This is the backwards question. It might be answered something like this:

  • To work out the number of triangles, Camilla subtracts one to use as the start then divides what's left by two to calculate the number of pairs of matches she can make.
  • T = (M - 1) 2.
  • If M = 37
    T = (37 - 1) 2
         = 36 2
         = 18
    She can make 18 triangles.

 

Keith sees the chain as beginning with one complete triangle of three matches. He then continues to make triangles by adding pairs of matches as far as required. He will add one less pair than the number of triangles. Say it, write the words then change to symbols, something like:
  • M = 3 + (T - 1) x 2
  • If T = 32
    M = 3 + (32 - 1) x 2
         = 3 + 31 x 2
         = 3 + 62 = 65

This is the backwards question. It might be answered something like this:

  • To work out the number of triangles, Keith subtracts three to use as the start then divides what's left by two to calculate the number of pairs of matches he can make. But he has to add in one to include his starting triangle.
  • T = (M - 3) 2 + 1.
  • If M = 49
    T = (49 - 3) 2 + 1
         = 46 2 + 1
         = 23 + 1
         = 24
    He can make 24 triangles.

 

Mia imagines the chain made from full triangles. That way she knows to multiply by three. Then she takes out one (the red one) at all the joins. There will be one less to take out than the number of triangles. Say it, write the words then change to symbols, something like:
  • M = 3 x T - (T - 1)
  • If T = 51
    M = 3 x 51 - (51 - 1)
         = 153 - 50
         = 103

This backwards question is very likely to force Mia to rethink her method. As natural as it may be, it is not clear how it can be used to tackle the inverse problem. Mia might need to find an equivalent to her equation in order to calculate plants from tiles. This may be an opportunity to suggest simplifying her equation:
M = 3 x T - (T - 1)
     = 3T - T + 1
     = 2T + 1
Which is simple for both substitution (as above for 51 triangles) and for solution.
If M = 107, the equivalent equation to Mia's rule shows the M = 53.

This example supports a case for valuing the simplest algebraic form of an expression.

 

Waleed sees the triangle chain as made of point up and point down triangles. But he also realises that the point down ones only happen because a match has been placed between each pair of point up triangles (the red ones) - like a bridge between the peaks of two mountains.

Then he realises that if the total number of triangles is even:

  • the number of red ones is the same as the number of triangles, and
  • he needs one extra match (the blue one) to complete the last triangle.
But, if the total number of triangles is odd:
  • the number of red ones is one less than the number of triangles.
For an even total Waleed needs:
  • 3 matches for each point up triangle
  • 1 match between each of these
  • 1 extra match.
But how many triangles are point up? Half of the total and that's the same number for the bridging matches.

So, M = 3 x (T/2) + T/2 + 1

If T = 26
M = 3 x (26/2) + 26/2 + 1
     = 3 x 13 + 13 + 1
     = 39 + 13 + 1
     = 53

For an odd total Waleed needs:
  • 3 matches for each point up triangle
  • 1 match between each of these
Halving the total gives a 'half number' this time, so how many triangles are point up and how many bridging matches are there? Use the whole number above the 'half number' for the point up triangles and the whole number below that for the bridging matches.

So, M = 3 x (T/2 + 1/2) + (T/2 - 1/2)

If T = 71
M = 3 x (71/2 + 1/2) + (71/2 - 1/2)
     = 3 x 72/2 + 70/2
     = 3 x 36 + 35
     = 108 + 35
     = 143

This backwards question is very likely to force Waleed to rethink his method. As creative has his method is, it is not clear how it can be used to tackle the inverse problem. Waleed might need to find an equivalent to his equations in order to calculate triangles from matches. This may be an opportunity to suggest simplifying his equations.

It is satisfying to discover that both his equations are equivalent to M = 2T + 1, the simplest algebraic expression governing this investigation. Try it. From that equation he can work backwards and discover that 57 matches will make 28 peaks.

To learn more about this Picture Puzzle click the photo to visit the Task Cameo.

Summary

Matches are arranged as chain of overlapping mountain peaks which make hidden or unseen triangles. First students are asked to find the total number of matches given small numbers of peaks, then calculate the number matches for 100 peaks. This leads into the general question If I tell you any number of peaks, can you tell me the total number of matches needed to make them?.

A mathematician asks Can I check it another way? and later in this Picture Puzzle Ben, Dani and Adrian reveal their ways of seeing the pattern. Students are then invited to explain and calculate in these ways, one of which might be the one they were already using.

Answers


Answers will vary because the students are making their own chains. However, you could encourage them to record peak and match numbers.

The process of making the peaks encourages the brain to predict the number of matches needed for other cases. This slide is only asking for an oral response but...

...this next one expects that to be written. The symbols don't have to be used, but since they are there, why not? ... as long as learners attribute meaning to them.
  • For 100 triangles the number of matches is 402, but ways of explaining will vary.
  • Different ways of seeing this experiment are the focus of the remainder of the puzzle.

 

Ben sees the pattern building from the hidden triangles, then he needs two matches at the end to finish the last peak that shows 'in front'. Encourage students to say, then write in words, then code the words as symbols. Something like:
  • M = P x 4 + 2
  • If P = 39
    M = 39 x 4 + 2
         = 156 + 2
         = 158

This is the backwards question. It might be answered something like this:

  • To work out the number of peaks Ben subtracts two to use at the end then divides what's left by four to calculate the number of peaks he can make.
  • P = (M - 2) 4.
  • If M = 54
    P = (54 - 2) 4
         = 52 4
         = 13
    He can make 13 peaks.

 

Dani builds the front peak first (the blue one) then the ones behind. The front peak needs six matches and the rest need four each. There is one less unseen triangle than the number of peaks. Dani might write in symbols something like:
  • M = 6 + (P - 1) x 4
  • If P = 44
    M = 6 + (44 - 1) x 4
         = 6 + (43 x 4)
         = 6 + 172
         = 178

This is the backwards question. It might be answered something like this:

  • To work out the number of peaks Dani subtracts six to use for the front peak then divides what's left by four to calculate the number of peaks she can make. Then she has to add one for the front peak.
  • P = (M - 6) 4 + 1.
  • If M = 102
    P = (102 - 6) 4 + 1
         = 96 4 + 1
         = 24 + 1
         = 25
    She can make 25 peaks.

 

Adrian imagines every peak made from 6 matches. It's just that for all except the front one, two matches (the red ones) are hidden. So his way to calculate is to multiply the number of peaks by 6 then subtract 2 matches for each peak except the front one. He might write in symbols something like:

  • M = P x 6 - (P - 1) x 2
  • If P = 82
    M = 82 x 6 - (82 - 1) x 2
         = 492 - (81 x 2)
         = 492 + 162
         = 330
This backwards question is very likely to force Adrian to rethink his method. To work backwards he first had to add on one less than the number of peaks. But the number of peaks is what he is trying to find out. So, as creative has his method is, it is not clear how it can be used to tackle the inverse problem. Adrian might need to find an equivalent to his equation in order to calculate peaks from matches. This may be an opportunity to suggest simplifying his equations.

M = P x 6 - (P - 1) x 2
     = 6P - 2P + 2
     = 4P + 2
From this simpler form he can easily work backwards and discover that 38 matches will make 9 peaks.

 

To learn more about this Picture Puzzle click the photo to visit the Task Cameo.