Grade 4 · 4.OA.C.5

Number and shape patterns from a rule

Tejas starts at 7 and keeps adding 3. The rule says nothing about odd and even numbers, so why do his numbers take turns, odd, even, odd, even, forever?

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When you reach for it

Tejas writes a list of numbers. He starts at 7, and each new number is 3 more than the one before:

7,\ 10,\ 13,\ 16,\ 19,\ \ldots

A list like this is a number pattern. Each number in it is a term: 7 is the first term, 10 the second term, 13 the third. The words “start at 7, add 3” are the pattern’s rule. The rule tells you how to make the pattern, one term at a time, for as long as you like.

Figure 1The pattern “start at 7, add 3”. Every hop is the same length, so the terms sit evenly spaced along the line: 7, 10, 13, 16, and so on.

Reach for a rule whenever something grows (or shrinks) by the same amount again and again: a savings jar that gets 3 more dollars each week, a stack that loses 8 books at each trip, a shape that gets 3 more sticks of wood at each step. The rule builds the list. Then comes the more interesting part. Look hard at the list, and it shows things the rule never said. Tejas’s terms take turns being odd and even. His rule says nothing about odd or even. A good question is always: why does that happen?

The procedure

Part 1 (day 1): building a pattern from its rule

A rule has two pieces: where to start and what to do each time. “Start at 7, add 3” gives the first term, 7, and the step that turns any term into the next one. Both pieces matter. “Add 3” alone could start anywhere: from 7 it gives 7, 10, 13, \ldots and from 20 it gives 20, 23, 26, \ldots, a different pattern with no terms in common with the first.

Each term comes from the one before it. To get the 5th term you need the 4th, and to get that you need the 3rd. So a pattern is built in order, and one slip spoils every term after it. A quick check at the end catches most slips (Example 1 shows one).

Rules can take away as well as add. “Start at 90, subtract 8” makes the terms smaller each time. With whole numbers (the numbers 0, 1, 2, 3, and so on) the pattern has to stop before it would go below 0.

Counting the steps. Going from the 1st term to the 5th term takes 4 steps, not 5. Each step adds one 3, so

\text{5th term} = 7 + 3 + 3 + 3 + 3 = 7 + 4 \times 3 = 19

In plain words, the 5th term is 7 plus four threes. The same counting says the 10th term is 7 plus nine threes. This lets you check a term, or jump ahead, without writing every term.

Is a number in the pattern? Every term is the start plus some number of threes. So a number is a term of Tejas’s pattern exactly when taking 7 away from it leaves a multiple of 3 (or leaves 0, for the first term itself). This turns a question about a long list into one division.

Part 2 (day 2): features the rule does not say

Once a pattern is built, look at it for things the rule never stated: are the terms odd or even, does the ones digit repeat, does every term share a factor? Noticing is the first half. The second half is explaining why, using the rule. A feature you can explain is a feature you can trust for every term, even the ones you never wrote down.

Odd and even when you add. Every whole number is even (it can be split into pairs with none left over) or odd (one is left over). The last digit tells which, as Divisibility rules showed. When you add two numbers:

first number	number added	sum
even	even	even
even	odd	odd
odd	even	odd
odd	odd	even

The table is about pairs. Two even numbers are both made of pairs, so their sum is too. Two odd numbers each have one left over, and the two leftovers make a new pair, so the sum is even. An odd and an even number together still have exactly one left over, so the sum is odd.

Using the table on a rule. Tejas adds 3, an odd number, at every step. So every step flips the term from odd to even or from even to odd. He starts at 7, which is odd, so the terms go odd, even, odd, even, for as long as the pattern runs. If a rule adds an even number instead, every step keeps the term the same kind: start odd and all the terms are odd; start even and all are even.

The ones digit. Adding 10 never changes the ones digit, and adding 20, 30 or 40 does not either. So if a few steps together add up to a multiple of 10, the ones digits come back to where they were and repeat from there. With “add 4”, five steps add 5 \times 4 = 20, so the ones digits repeat every five terms.

Part 3 (day 3): shape patterns and multiplying rules

A shape can follow a rule. Eydis builds squares in a row from wooden dowels. One square needs 4 dowels. To add a second square, she does not need 4 more, because the new square shares a side with the old one: she adds 3 dowels, a top, a bottom and a new right side. Every new square is the same, so the dowel counts make the pattern “start at 4, add 3”.

Figure 2Eydis’s row of squares. Each new square shares one side with the square before it, so it needs only 3 new dowels: 4, then 7, then 10.

The number of squares is the step number. Step 1 has 1 square, step 2 has 2 squares, and so on. Turning the picture into a rule is the work; once you have the rule, everything from Parts 1 and 2 applies, including jumping to a far-off step without drawing every shape.

A shape often has a second way to count. For the row of squares, start with the single dowel on the far left, and then give each square its own 3 dowels: top, bottom and right side. That counts the dowels for 20 squares as 1 plus twenty threes, while the rule counts them as 4 plus nineteen threes. Two counts that agree are a strong check.

Multiplying rules. A rule can multiply instead of add: “start at 4, multiply by 3”. Each term is 3 times the one before. Multiplying by 3 makes the terms grow much faster than adding 3 does: each step adds two copies of the term it starts from, and the terms keep getting bigger, so each step adds more than the step before. The same habit of looking for features applies. Products follow their own odd and even facts: a product is odd only when every number multiplied is odd. If even one of the numbers is even, the product is even, since it is made of whole pairs.

Worked examples

Examples 1, 2 and 3 go with Part 1, Examples 4 and 5 with Part 2, and Examples 6 and 7 with Part 3.

Example 1

Building Tejas’s pattern

Use the rule “start at 7, add 3” to write the first 8 terms.

Solution

The first term is the start, 7. Each new term is 3 more than the term before, because that is the step the rule gives:

7,\ 10,\ 13,\ 16,\ 19,\ 22,\ 25,\ 28

Check the last term by counting steps instead of adding one at a time. From the 1st term to the 8th is 7 steps, so the 8th term is 7 plus seven threes: 7 + 7 \times 3 = 7 + 21 = 28. It matches.

Example 2

A pattern that runs down

Ameyali has a stack of 90 flyers. On each trip down the street she hands out 8. Write the number of flyers left after each trip, starting with the 90 before she leaves, for as long as she can make a full trip. How many flyers are left at the end?

Solution

The rule is “start at 90, subtract 8”, since each trip takes away the same 8 flyers:

90,\ 82,\ 74,\ 66,\ 58,\ 50,\ 42,\ 34,\ 26,\ 18,\ 10,\ 2

After the term 2 she cannot hand out 8 more, since 2 - 8 would be less than 0. So the pattern stops at 2, and 2 flyers are left. The list has 12 terms, the first one being the start, so she made 11 full trips. Check: 11 trips of 8 is 88 flyers, and 90 - 88 = 2.

Example 3

Is it in the pattern?

Is 52 a term of Tejas’s pattern “start at 7, add 3”? Is 60? If a number is a term, say which term it is.

Solution

Every term is 7 plus some threes. So take away the start and see whether what is left is a multiple of 3.

For 52: 52 - 7 = 45, and 45 \div 3 = 15 \text{ R } 0. So 52 is 7 plus fifteen threes. That is 15 steps past the first term, which makes 52 the 16th term.

For 60: 60 - 7 = 53, and

53 \div 3 = 17 \text{ R } 2

The remainder is not 0, so 60 is not a term. Seventeen threes after 7 reach 7 + 17 \times 3 = 58, and one more step jumps to 61. The number 60 falls in between and is never landed on.

Example 4

Odd, even, odd, even

Look at the first 8 terms of Tejas’s pattern from Example 1. Which are odd and which are even? Explain why this keeps going, and decide whether the 15th term is odd or even without finding it.

Solution

The terms are 7, 10, 13, 16, 19, 22, 25, 28. By their last digits, 7, 13, 19, 25 are odd and 10, 16, 22, 28 are even. They take turns.

The rule never mentions odd or even, so the reason has to come from the step. Each step adds 3, which is odd. By the table in Part 2, odd plus odd is even and even plus odd is odd. So every step flips the kind of number, and the flipping never stops. The first term, 7, is odd, so the 1st, 3rd, 5th, 7th terms are odd: every term in an odd-numbered place. The 15th term is in an odd-numbered place, so it is odd.

Check by counting steps: the 15th term is 7 plus fourteen threes, 7 + 14 \times 3 = 7 + 42 = 49, and 49 is odd.

Example 5

Eydis’s ones digits

Eydis uses the rule “start at 3, add 4”. Write the first 11 terms. Then describe two features the rule does not say, and explain each one.

Solution

Start at 3 and add 4 each time:

3,\ 7,\ 11,\ 15,\ 19,\ 23,\ 27,\ 31,\ 35,\ 39,\ 43

First feature: every term is odd. The start, 3, is odd. Each step adds 4, which is even, and odd plus even is odd. So an odd term always leads to another odd term, and every term is odd.

Second feature: the ones digits repeat. The ones digits are 3, 7, 1, 5, 9, then 3, 7, 1, 5, 9 again, then 3. Five steps add 5 \times 4 = 20 to a term, and adding 20 changes only the tens digit. So every term has the same ones digit as the term five places before it: 23 and 3, 27 and 7, 31 and 11. The ones digits repeat in a cycle of five, forever. The cycle 3, 7, 1, 5, 9 uses every odd digit exactly once, and that has a reason too. Every term is odd, so every ones digit is odd. Inside one cycle no digit comes back early, since 1, 2, 3 or 4 steps add 4, 8, 12 or 16, and none of those ends in 0. Five different odd digits are all five odd digits.

Example 6

Twenty squares

Eydis keeps building her row of squares from dowels (the figure in Part 3). How many dowels does a row of 20 squares need? And which step uses exactly 100 dowels?

Solution

The dowel counts follow the rule “start at 4, add 3”: step 1 needs 4, and each new square adds 3. Step 20 is 19 steps past step 1, so it needs 4 plus nineteen threes:

4 + 19 \times 3 = 4 + 57 = 61

Check with the second way to count from Part 3: one dowel on the far left, plus 3 for each of the 20 squares, gives 1 + 20 \times 3 = 1 + 60 = 61. The two counts agree, so 20 squares need 61 dowels.

For 100 dowels, take away the start: 100 - 4 = 96, and 96 \div 3 = 32 \text{ R } 0. So 100 is 4 plus thirty-two threes, which is 32 steps past step 1. Step 33 uses exactly 100 dowels. Check: 1 + 33 \times 3 = 1 + 99 = 100.

Example 7

A multiplying rule

Tejas tries the rule “start at 4, multiply by 3”. Write the first 6 terms. Then explain why every term is even, and why every term is a multiple of 4.

Solution

Each term is 3 times the term before:

4,\ 12,\ 36,\ 108,\ 324,\ 972

For example, 3 \times 36 = 108 and 3 \times 108 = 324.

Every term is even. The first term, 4, is even. Each step multiplies an even number by 3, and a product with an even number in it is even, since 3 groups of an even number are all made of pairs. So an even term always leads to another even term.

Every term is a multiple of 4. The 2nd term is 4 \times 3. The 3rd is 4 \times 3 \times 3, and the 6th is 4 times five threes multiplied together. Each one is 4 times a whole number, which is what a multiple of 4 is. Check one: 972 \div 4 = 243 \text{ R } 0.

Compare this with “start at 4, add 3” from Example 6, which reached only 19 by its 6th term. Multiplying by 3 adds more at every step than the step before did (8, then 24, then 72), so it races ahead.

Boundaries

A rule needs a start. “Add 3” with no starting number does not fix a pattern; each start gives a different one.
Subtracting stops at 0. With whole numbers, a rule that subtracts runs only until the next term would be less than 0. “Start at 90, subtract 8” ends at 2.
A feature is a guess until the rule explains it. The pattern “start at 12, add 11” begins 12, 23, 34, 45, 56, 67, 78, 89, and in each term the ones digit is one more than the tens digit. The next term is 100, and that feature is gone. Only a reason drawn from the rule makes a feature safe for every term.
The first few terms do not fix the rule. The list 1, 2, 4 could come from “multiply by 2” (next term 8) or from “add 1, then add 2, then add 3” (next term 7). That is why this lesson always starts from a given rule.
Steps, not terms. The 10th term is the start plus nine steps, not ten. The first term is the start itself and has had no steps.

Where this goes next

A rule that makes each term from the one before is the seed of Recursive formulas, where the same idea is written with symbols. A pattern can also be laid beside a second pattern, term by term, and the pairs graphed to see how the two patterns are related: that is Two number patterns side by side and graphing their pairs.

Exercises

Exercises 1 to 8 go with Part 1, 9 to 15 with Part 2, and 16 to 21 with Part 3. The reasoning group draws on all three.

Part 1: building a pattern from its rule

Write the first 6 terms of “start at 2, add 5”.
Write the first 6 terms of “start at 11, add 9”.
Write the first 8 terms of “start at 100, subtract 7”.
Write the first 5 terms of “start at 15, add 12”.
Ameyali has 60 stickers and gives 7 to each friend in turn, for as long as she can give a full 7. Write the number of stickers she has before and after each friend. How many friends get stickers, and how many stickers are left?
Is 47 a term of “start at 5, add 7”? If so, which term?
Is 80 a term of “start at 2, add 4”? If not, name the terms just below and just above it.
A pattern has the rule “add 9”. Its first five terms are 14, \square, 32, \square, 50. Fill in the two missing terms.

Part 2: features the rule does not say

Write the first 6 terms of “start at 4, add 5”. Describe what happens to odd and even and to the ones digits, and explain why.
Write the first 6 terms of “start at 1, add 10”. What stays the same in every term, and why?
Write the first 5 terms of “start at 8, add 2” and of “start at 9, add 2”. One pattern is all even and one is all odd. Explain why.
Without listing the terms, decide whether the 20th term of “start at 6, add 5” is odd or even. Then check by finding it.
Write the first 6 terms of “start at 1, add 8”. Which ones digits appear, and after how many terms do they repeat? Explain the repeat.
Is any term of Tejas’s pattern “start at 7, add 3” a multiple of 3? Explain.
Write the pattern “start at 50, subtract 5” until it cannot go on. How many terms are there, and what do all the terms have in common?

Part 3: shape patterns and multiplying rules

Figure 3Ameyali’s L of coasters, for Exercises 16 and 17. Each step adds one coaster to the top of the L and one to the end of its foot.

Ameyali builds an L of coasters (the figure below). Write the rule for the number of coasters, and find how many coasters step 6 needs.
In Ameyali’s L of coasters, which step uses exactly 25 coasters? Could any step use exactly 30?
Eydis builds a new row of squares from dowels, this time with a diagonal dowel across each square. One square needs 5 dowels, and each new square adds 4: a top, a bottom, a right side and a diagonal. How many dowels do 12 squares need?
Write the first 6 terms of “start at 5, multiply by 3”. Explain why every term is odd and why every term ends in 5.
Write the first 6 terms of “start at 7, multiply by 2”. Which terms are odd, and why?
Compare “start at 3, add 3” with “start at 3, multiply by 3”. Write the first 6 terms of each. Which term of each pattern is the first one bigger than 100?

Reasoning

A rule starts at an even number and adds an odd number. Explain why its terms must take turns, even, odd, even, odd.
Tejas says, “My rule is start at 7, add 3, so my 10th term is 7 + 10 \times 3 = 37.” Is he right? Explain and give the correct 10th term.
Ameyali says the pattern “start at 10, subtract 3” goes on forever, the way adding does. Is she right for whole numbers? Explain.
Eydis says a row of 5 squares made of dowels needs 5 \times 4 = 20 dowels, because each square has 4 sides. Explain what she has missed, and find the right number.

Answers

2, 7, 12, 17, 22, 27. Add 5 each time; check: 2 + 5 \times 5 = 27.
11, 20, 29, 38, 47, 56. Check: 11 + 5 \times 9 = 56.
100, 93, 86, 79, 72, 65, 58, 51. Check: 100 - 7 \times 7 = 51.
15, 27, 39, 51, 63. Check: 15 + 4 \times 12 = 63.
60, 53, 46, 39, 32, 25, 18, 11, 4. After 4 she cannot give a full 7, so 8 friends get stickers and 4 are left. Check: 8 \times 7 = 56 and 60 - 56 = 4.
Yes, the 7th term. 47 - 5 = 42 = 7 \times 6, so 47 is 5 plus six sevens, 6 steps past the first term.
No. 80 - 2 = 78 and 78 \div 4 = 19 \text{ R } 2. The terms around it are 2 + 19 \times 4 = 78 and 82.
23 and 41. 14 + 9 = 23 and 32 + 9 = 41; check: 23 + 9 = 32 and 41 + 9 = 50.
4, 9, 14, 19, 24, 29. They take turns even, odd, even, odd, because adding the odd number 5 flips odd and even each step. The ones digits take turns 4 and 9, because two steps add 2 \times 5 = 10, which leaves the ones digit unchanged.
1, 11, 21, 31, 41, 51. The ones digit is always 1, because adding 10 changes only the tens digit (each step adds one ten).
8, 10, 12, 14, 16 and 9, 11, 13, 15, 17. Adding 2, an even number, never changes odd or even: even plus even is even, odd plus even is odd. So the first pattern, starting even, stays even, and the second, starting odd, stays odd.
Odd. The start, 6, is even, and each step adds the odd number 5, so the terms go even, odd, even, odd. Terms in even-numbered places are odd, and 20 is even. Check: the 20th term is 6 + 19 \times 5 = 101, which is odd.
1, 9, 17, 25, 33, 41. The ones digits are 1, 9, 7, 5, 3, and then 1 again at the 6th term, so they repeat every 5 terms. Five steps add 5 \times 8 = 40, and adding 40 leaves the ones digit unchanged.
No. The first term is 7 = 2 \times 3 + 1, one more than a multiple of 3. Each step adds a whole 3, so every term stays one more than a multiple of 3: 10 = 9 + 1, 13 = 12 + 1. Dividing any term by 3 leaves remainder 1, never 0.
50, 45, 40, 35, 30, 25, 20, 15, 10, 5, 0, which is 11 terms; the next would be less than 0. Every term is a multiple of 5 (the start is, and taking away 5 keeps it so), and the ones digits take turns 0 and 5: one step takes away 5, and two steps take away 10, which leaves the ones digit where it was.
“Start at 3, add 2”: step 1 has 3 coasters, and each step adds one at the top and one at the end of the foot. Step 6 needs 3 + 5 \times 2 = 13 coasters.
Step 12: 25 - 3 = 22 and 22 \div 2 = 11, so 25 is 3 plus eleven twos, 11 steps past step 1. No step uses 30: the start is odd and each step adds the even number 2, so every count is odd.
49 dowels. The rule is “start at 5, add 4”, and 12 squares is 11 steps past one square: 5 + 11 \times 4 = 49. Check: 1 dowel on the far left plus 4 for each square gives 1 + 12 \times 4 = 49.
5, 15, 45, 135, 405, 1215. Each term is 3 times an odd number, and a product of odd numbers is odd, so every term is odd. The ones digit of a product comes only from the ones digits multiplied (the tens and hundreds make only tens and hundreds). A number ending in 5 times 3 therefore ends like 5 \times 3 = 15, in 5, so the ending 5 is passed on at every step.
7, 14, 28, 56, 112, 224. Only the first term, 7, is odd. The 2nd term is 7 \times 2, which has a factor of 2, and every term after it is an even number times 2. A product with an even number in it is even.
3, 6, 9, 12, 15, 18 and 3, 9, 27, 81, 243, 729. The multiplying pattern first passes 100 at its 5th term, 243. The adding pattern is at 99 = 3 + 32 \times 3 at its 33rd term, so its first term past 100 is the 34th, 102.
The first term is even. Each step adds an odd number. Even plus odd is odd, and odd plus odd is even, so each step flips the kind of number. Starting from even, the flips give even, odd, even, odd, with no end.
No. The 1st term is 7 with no threes added, so the 10th term has only 9 steps after it. The 10th term is 7 plus nine threes: 7 + 9 \times 3 = 34.
No. The terms are 10, 7, 4, 1. The next would be 1 - 3, less than 0, which is not a whole number, so the pattern stops at 1 after 4 terms. Adding can always go on; subtracting runs out.
Neighboring squares share a side, so counting 4 for every square counts each shared dowel twice. Five squares in a row share 4 sides, so the right number is 20 - 4 = 16. The rule “start at 4, add 3” agrees: 4 + 4 \times 3 = 16.