13.4Digit Budget Navigator
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Chapter 13 • Page Number Problems

Work Backward from a Digit Budget

第13讲 · 从数字预算倒推页数

A book gives you the total number of digit symbols used to print all page numbers. Your job is to spend that budget on complete one-, two-, three-, and four-digit blocks, then identify the final page—or prove that the budget is impossible.

Spend complete blocks first. Divide only inside the correct block.

Grade 510 missionsSelf-containedAbout 45–60 minutes
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Mission 1

Build the digit-budget staircase

Before working backward, memorize nothing. Rebuild the complete blocks from place value.

Not complete
Pages 1–9

9 page labels × 1 digit

9 digits
Cumulative: 9
Pages 10–99

90 page labels × 2 digits

180 digits
Cumulative: 189
Pages 100–999

900 page labels × 3 digits

2,700 digits
Cumulative: 2,889
Pages 1000–9999

9,000 page labels × 4 digits

36,000 digits
Cumulative: 38,889
Work backward by removing every complete lower-width block before dividing the leftover budget.
Keep the units visible. Here “budget” means the exact number of digit symbols used, not an allowance you may leave partly unused. It counts digit symbols, not pages. A three-digit page costs exactly 3 units from the digit budget.

Check the complete-block totals

Hint 1 — choose a starting point

Rebuild the full blocks from label counts and widths.

Hint 2 — take the next step

Do not divide the whole budget by the final width.

Mission 2

Follow the worked example’s 723-digit solution

The worked example chooses page 200 as a checkpoint, spends the known budget through that page, and then continues.

Not complete

Complete practice problem: Printing all page numbers of a book uses 723 digit symbols. How many pages does the book have?

Pages 1–99 one-digit page labels9 digits
Pages 10–9990 two-digit page labels180 digits
Pages 100–200200 − 100 + 1 = 101 three-digit page labels303 digits
Budget left723 − 9 − 180 − 303231 digits
Continue after page 200231 ÷ 377 pages
9180303231
200 + 77 = 277 pages

Reconstruct the worked example calculation

Hint 1 — choose a starting point

This worked solution uses page 200 as a checkpoint.

Hint 2 — take the next step

Subtract the checkpoint cost; each later label costs three digits.

Mission 3

Use the general block method—and avoid an off-by-one error

A more reusable route removes the complete one- and two-digit blocks, then counts how many three-digit page labels the leftover buys.

Not complete

Spend the complete lower blocks

723 − 189 = 534 digits

The remaining budget is entirely inside the three-digit block.

534 ÷ 3 = 178 page labels

Convert a count into an endpoint

First page + count − 1
100 + 178 − 1 = 277

We subtract 1 because page 100 is already the first of the 178 labels.

First label
100
+ 178 − 1 →
Last label
277

Endpoint mini-laboratory

100 + 178 − 1 = 277

Audit the general method

Hint 1 — choose a starting point

Remove 189 before dividing by three.

Hint 2 — take the next step

The endpoint is first label + number of labels − 1.

Mission 4

Guided practice: a dictionary uses 2,925 digits

This practice crosses the 999/1000 boundary, so the remaining page labels are four digits wide.

Not complete

Complete practice problem: Printing the page numbers of a student dictionary uses 2,925 digit symbols. How many pages does the dictionary have?

Pages 1–9999 + 180 + 2,7002,889 digits
Budget left2,925 − 2,88936 digits
Four-digit labels36 ÷ 49 pages
100010011002100310041005100610071008
1000 + 9 − 1 = 1008 pages

Complete the guided practice

Hint 1 — choose a starting point

Use the complete total through page 999.

Hint 2 — take the next step

The next labels have four digits.

Mission 5

Use an inverse digit-budget laboratory

Enter an exact whole-number digit total from 1 through 200,000. The laboratory spends complete blocks, checks divisibility inside the final block, and reports the endpoint or an impossibility.

Not complete
Page 1,994

Exactly 6,869 digit symbols are used.

BlockAvailable costBudget spentBudget left
Feasibility rule: once the budget reaches its final digit-width block, the leftover must be divisible by that width.

Use the laboratory

Hint 1 — choose a starting point

Spend lower-width blocks first.

Hint 2 — take the next step

The leftover must buy a whole number of labels.

Mission 6

Audit boundary budgets and impossible leftovers

At 9/10, 99/100, and 999/1000, each new page label has a wider cost.

Not complete
9 → 10
9→11
+2 digits
99 → 100
189→192
+3 digits
999 → 1000
2,889→2,893
+4 digits

Why 724 is impossible

724 − 189 = 535
535 = 178 × 3 + 1

After 178 complete three-digit labels, one digit remains. That would stop partway through page 278.

Nearby possible budgets

720 → page 276
723 → page 277
726 → page 278

Inside the three-digit block, attainable budgets rise in steps of 3.

Check the boundary endpoints

Hint 1 — choose a starting point

Look immediately below and above the boundary.

Hint 2 — take the next step

A leftover digit cannot become a partial page label.

Mission 7

Share one digit budget between two books

Each book numbers its own pages from 1, so the shorter labels occur separately in both books. Two examples give a combined digit budget and a difference in page counts. Remove the cost of the extra pages first.

Not complete

original exercise A

Establish the width first. If the smaller book had at most 99 pages, the larger would have at most 104; together they would use at most 189 + 204 = 393 digits. That is too few. If the larger reached 1000, even the smaller would reach 995, and their combined budget would exceed 5700. Both endpoints therefore have three digits.

Two volumes use 687 digit symbols altogether. The upper volume has 5 more pages than the lower volume. How many pages does the upper volume have?

5 extra three-digit pages cost 5 × 3 = 15 digits
687 − 15 = 672; 672 ÷ 2 = 336 digits per equalized book
336 − 189 = 147; 147 ÷ 3 = 49; 99 + 49 = 148
Lower volume: 148 pages • Upper volume: 153 pages. Check: 336 + 351 = 687 digits, and 153 − 148 = 5 pages.

original exercise B

The same boundary check applies: endpoints at most 99 and 106 cost at most 399 digits, below 777. A four-digit endpoint would push both books near 1000 and far above 777. Both endpoints are three-digit.

Books A and B use 777 digit symbols altogether. Book A has 7 more pages than Book B. How many pages does Book A have?

7 × 3 = 21 extra digits
777 − 21 = 756; 756 ÷ 2 = 378 digits for Book B
Book B: 162 pages • Book A: 169 pages. Check: 378 + 399 = 777 digits, and 169 − 162 = 7 pages.

Two-book solver

Use an exact combined total from 1–200,000 and a whole-number page difference from 0–99,999. A difference of zero means equal-length books. Each book has at least one page.

Across a boundary, do not assume every extra page has three digits. Books with 98 and 101 pages differ by three pages, but the extra labels 99, 100, 101 cost 2 + 3 + 3 = 8 digits, not 9.

Smaller: 148 pages • Larger: 153 pages

Complete both original exercises

Hint 1 — choose a starting point

Establish the digit width of both books before equalizing them.

Hint 2 — take the next step

Subtract the digit cost of the extra labels, then halve. Check both original conditions.

Mission 8

Use the same budget idea inside 123456789101112…

Instead of asking for a final page, this extension asks which number contains a distant digit—and which digit inside that number is needed.

Not complete

Counting-string extension: In the decimal 0.1234567891011121314…, what is the 2,016th digit after the decimal point?

One-digit numbers 1–99 × 19 positions
Two-digit numbers 10–9990 × 2180 positions
Position left inside three-digit numbers2,016 − 1891,827
Three-digit number index1,827 ÷ 3609th

Find the containing number

100 + 609 − 1 = 708

The 609th three-digit number is 708.

Choose the digit inside it

1,827 is divisible by 3

So the target is the final digit of 708:

8
Digit 8

Position 2,016 is the third digit of 708.

When the position is not a multiple of the width

Position 193 leaves 193 − 189 = 4 positions. Remove the three digits of 100: one remains, so this is the first digit of 101, namely 1. A remainder of two selects the second digit; a remainder of zero selects the last digit of the preceding complete number.

Nearby boundary positions

The highlighted box is the selected position. The small label underneath is its overall position.

Locate the 2,016th digit

Hint 1 — choose a starting point

Subtract all positions occupied by shorter numbers.

Hint 2 — take the next step

Find the containing number and then the position inside it.

Mission 9

Independent workshop

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

Not complete
Hint

Remove complete lower-width blocks before dividing.

Worked explanation — open after trying

Subtract 9, divide by 2, then add 9: 88 pages.

Hint

Remove complete lower-width blocks before dividing.

Worked explanation — open after trying

Subtract 189, divide by 3, then add 99: 145 pages.

Hint

Remove complete lower-width blocks before dividing.

Worked explanation — open after trying

Subtract 189, divide by 3, then add 99: 306 pages.

Hint

Remove complete lower-width blocks before dividing.

Worked explanation — open after trying

Subtract 2889, divide by 4, then add 999: 1003 pages.

Hint

Remove complete lower-width blocks before dividing.

Worked explanation — open after trying

Subtract 2889, divide by 4, then add 999: 1205 pages.

Hint

Compare the totals through 99 and 100.

Worked explanation — open after trying

Page 99 ends at 189 digits and page 100 ends at 192. There is no complete endpoint at 190.

Hint

Check the digit widths before equalizing the books.

Worked explanation — open after trying

Both books must have three-digit endpoints: even 99 and 103 use only 390 digits; endpoints near 1000 use far more than 552. Remove 4 × 3 = 12, then halve: 270 digits. 99 + (270 − 189) ÷ 3 = 126 for the smaller; larger 130. Check: 270 + 282 = 552.

Hint

Remove 189 positions, then locate the position inside a three-digit number.

Worked explanation — open after trying

195 − 189 = 6. The sixth digit in 100101… is the last digit of 101, which is 1.

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 final block uses three-digit labels.

Worked explanation — open after trying

99 + (594 − 189) ÷ 3 = 234.

Hint

The total through 999 is 2889.

Worked explanation — open after trying

999 + (2917 − 2889) ÷ 4 = 1006.

Hint

Compare the complete digit totals on either side of the next width boundary.

Worked explanation — open after trying

The neighboring totals are 189 and 192, so no.

Hint

Use the combined budget to establish both endpoint widths before pricing the six extra pages.

Worked explanation — open after trying

If the smaller were at most 99, the total would be at most D(99)+D(105)=396. Four-digit endpoints would cost over 5700. Remove 18 and halve to get 228 digits. Smaller: 99 + (228−189)÷3=112; larger 118. Check 228+246=474.

Hint

Find the digit’s position inside its number, not just the number.

Worked explanation — open after trying

194 − 189 = 5. In 100101…, the fifth digit is the middle digit of 101: 0.

Lesson checkpoints completed

Digit-Budget Reverse Engineer

This certifies that Learner can spend complete page-number blocks, work backward to a final page, detect impossible budgets, and locate distant digits in a concatenated counting string.

Lesson 13.4 • Grade 5 Math

Optional reflection — not automatically graded

The 2,016th-digit locator is a clearly labeled extension from the corresponding Chapter 13 Practice .