13.6Math Explorer
0 of 10 missionsChapter 13
Grade 5 • Chapter 13 • Lesson 13.6

Use Facing Pages, Sheets, and Odd–Even Structure

用相对页、纸张与奇偶结构解决页码问题

Read the physical structure of a book: which pages face each other, which two labels lie on one leaf, and how an article’s length changes the parity of the next starting page. Then use those rules to prove a maximum.

A page number has a value—and a position in the book.
10 missionsOpen-book modelsParity studioEverything needed is on this page.
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Mission 1

Separate three book structures

“Consecutive,” “facing,” and “on the same physical leaf” do not mean the same thing.

Not complete
📖
Facing spread

Two pages visible at once.

2 & 3, 4 & 5, 6 & 7, …
even left • odd right
📄
One physical leaf

The front and back of one piece of paper.

1 & 2, 3 & 4, 5 & 6, …
odd front • next even back
🔢
Consecutive labels

Two numbers differ by one, but structure still matters.

22 & 23 are facing—not one leaf
check the smaller number
Convention used in this lesson: page numbering begins with page 1. In a normal left-to-right book, an open spread has an even page on the left and the next odd page on the right.

Classify the page relationships

Hint 1 — choose a starting point

Decide whether the object is a spread or one leaf.

Hint 2 — take the next step

A spread is even/odd; one leaf is odd/even.

Mission 2

Explore facing pages

Under this lesson’s numbering convention, an open spread is determined by its even left page. The pair checker accepts the two page numbers in either order.

Not complete
42LEFT • EVEN
43RIGHT • ODD

Move through the book

Facing pages = n and n + 1, where n is even.

Pair checker

Read the facing-page rule

Hint 1 — choose a starting point

The facing labels differ by one.

Hint 2 — take the next step

The smaller label must be even.

Mission 3

original guided practice: facing pages with product 1,806

Translate “two facing pages” into two consecutive numbers with the smaller one even.

Not complete

Complete practice problem: A mathematics book is opened to two facing pages. The product of the page numbers is 1,806. Find the two page numbers.

Build the model

n(n + 1) = 1,806

Look for two neighboring factors. The smaller factor must be even.

42even left page
×
43odd right page

Verify every condition

42 × 43 = 1,806
43 − 42 = 1
42 is even
The pages are 42 and 43.

Consecutive-product explorer

42 × 43 = 1,806

Complete the worked example reasoning

Hint 1 — choose a starting point

Try neighboring factors near the square root.

Hint 2 — take the next step

Check product, adjacency, and even-left parity.

Mission 4

Track one physical leaf—and the middle leaf

The two sides of Leaf k carry page labels 2k − 1 and 2k. A “central leaf” means one leaf with equally many leaves before and after it; both sides of every leaf are numbered.

Not complete
Front • odd
47
Back • next even
48

Leaf-index laboratory

Leaf 24 carries pages 47 and 48.

Test 13 transfer: missing leaf sum 95

One physical leaf is missing, and its two page labels add to 95.

47 + 48 = 95

The front is page 47 and the back is page 48.

Test 13 transfer: central leaf product 2,450

49 × 50 = 2,450

Pages 49 and 50 are Leaf 25. If this is the central leaf, there are 49 leaves altogether, so the book has 98 numbered pages.

1 + ··· + 98 = 98 × 99 ÷ 2 = 4,851
Why 98 pages? Leaf 25 is central only when 24 leaves come before it and 24 come after it: 24 + 1 + 24 = 49 leaves. Each leaf has two numbered sides.

Use the leaf structure

Hint 1 — choose a starting point

Leaf k has labels 2k−1 and 2k.

Hint 2 — take the next step

A central leaf has the same number of leaves before and after it.

Mission 5

See how article length changes the next starting page

Parity means whether a number is odd or even. An article starting on page s and using L pages occupies s through s + L − 1, including both endpoints. The next article begins on s + L.

Not complete

Parity machine

Starting parity
Odd
+ even length
parity stays the same
Next parity
Odd
Starting parity
Odd
+ odd length
parity changes
Next parity
Even
Even length preserves parity. Odd length flips parity.

Apply the parity machine

Hint 1 — choose a starting point

The article occupies start through start + length − 1.

Hint 2 — take the next step

The next start is start + length.

Mission 6

Worked example: prove the maximum number of odd starting pages

Fifteen articles have lengths 1, 2, 3, …, 15 pages. They are bound back-to-back with continuous page numbering.

Not complete

Complete practice problem: Use each article length 1 through 15 exactly once, in any order. The first article starts on page 1; each next article starts immediately after the preceding article, with no blank pages. What is the greatest possible number of articles whose first page has an odd page number?

15articles altogether
8odd article lengths
4odd-length articles forced to start even
11largest possible odd starts

Why four odd lengths must start even

Each odd-length article flips the parity. Ignoring the even lengths between them, the eight odd-length articles start:

1st odd lengthodd start2ndeven start3rdodd start4theven start5thodd start6theven start7thodd start8theven start

Why 11 can actually be reached

The seven even-length articles do not change parity, so all seven can be placed during an odd-start phase. The worked example gives this order:

1, 3, 2, 4, 6, 8, 10, 12, 14, 5, 7, 9, 11, 13, 15

Only the articles of lengths 3, 7, 11, and 15 begin on even pages. The other 11 begin on odd pages.

A maximum proof needs two parts: show that more than 11 is impossible, and give an arrangement that reaches 11.

Rebuild the proof

Hint 1 — choose a starting point

Ignore even lengths temporarily and follow odd lengths.

Hint 2 — take the next step

Prove an upper bound, then construct an order that reaches it.

Mission 7

Build and audit an article arrangement

Each chip shows an article’s length in pages, not its starting page. Select a chip and move it earlier or later in the reading order. The table recalculates every starting page.

Not complete
8odd starts
7even starts
120total pages
PositionArticle lengthStarting pageStart parityNext start

Verify an optimal construction

Build your own order and check it together with the answers below. After an attempted check, you can open a separate worked example without changing your order. The Move buttons also work with a keyboard; the selected chip stays focused.

Mission completion also requires the current arrangement studio to show exactly 11 odd starts.

Hint 1 — choose a starting point

Put even-length articles where the start is odd.

Hint 2 — take the next step

Check the table after moving a chip; the current order must reach 11.

Mission 8

Transfer the structure to a February calendar

Find a 2 × 2 block whose four dates have written digits with the greatest possible sum.

Not complete

Test 13 transfer: Use the displayed 28-day February, whose first day is Saturday. A valid rectangle uses two neighboring weekday columns and two neighboring week rows, with a date in all four cells. It cannot include blank cells or wrap from Sunday to Monday. The example block 4, 5, 11, 12 has digit sum 14. What is the greatest possible digit sum?

Current block

4, 5, 11, 12
Digit sum = 14
1 of 17
Do not add the dates themselves. For 18, 19, 25, and 26, add the written digits: 1 + 8 + 1 + 9 + 2 + 5 + 2 + 6.

Find and verify the maximum

Hint 1 — choose a starting point

Add digits, not date values.

Hint 2 — take the next step

Visit every valid 2×2 block and compare its digit sum.

Mission 9

Independent workshop

Try all eight questions before opening the explanations. Correct all eight answers to complete this mission.

Not complete
Hint

A spread is even then next odd.

Worked explanation — open after trying

76 + 1 = 77.

Hint

One leaf has odd front, next even back.

Worked explanation — open after trying

85 + 1 = 86.

Hint

Try consecutive factors near the square root of 3192.

Worked explanation — open after trying

56 × 57 = 3192; 56 is even, so the facing pair is valid.

Hint

Next start = start + length.

Worked explanation — open after trying

25 + 14 = 39; parity is preserved.

Hint

The next start is one beyond the last occupied page.

Worked explanation — open after trying

25 + 13 = 38; parity flips.

Hint

Count odd lengths, then place all even lengths during an odd-start phase.

Worked explanation — open after trying

Five odd lengths alternate odd/even starts, giving three odd starts. All four even lengths can start odd: maximum 7. Order 2,4,6,8,1,3,5,7,9 reaches it.

Hint

Find the leaf number and count equal numbers of leaves on both sides.

Worked explanation — open after trying

This is Leaf 12. Eleven leaves on each side give 23 leaves, hence 46 pages.

Hint

Add individual digits.

Worked explanation — open after trying

1+7+1+8+2+4+2+5 = 30.

Mission 10

Fresh exit ticket

Use the method on new problems. All five answers must be correct. Complete all ten missions to earn the certificate.

Not complete
Hint

The smaller page must be even.

Worked explanation — open after trying

38 is even and 39 is the next odd page.

Hint

The odd front is one less.

Worked explanation — open after trying

64 − 1 = 63.

Hint

Add an even number to either an odd or even start.

Worked explanation — open after trying

An even length preserves parity.

Hint

Give both a bound and a construction.

Worked explanation — open after trying

Four odd lengths force two even starts. The three even lengths can all start odd. Maximum 5; order 2,4,6,1,3,5,7 reaches it.

Hint

Find the leaf’s position, then put equally many leaves before and after it.

Worked explanation — open after trying

15+1+15=31 leaves; 31×2=62 pages.

Chapter 13 Achievement

Book-Structure & Parity Navigator

This certifies that Learner can distinguish facing pages from one physical leaf, use even–odd page structure, solve consecutive-page products, and prove a maximum with parity and a construction.

Lesson 13.6 • Use Facing Pages, Sheets, and Odd–Even Structure
The interactive laboratories, extra practice, and assessments are new instructional scaffolds.