Problem
Count digits used over a range
Pages are numbered 1, 2, 3, … with none skipped. Printing all the page numbers took 81 digit characters. A one-digit page costs 1 character, a two-digit page costs 2. Find the number of the last page.
Reasoning · Grade 4-1 Numbers and Digit Counts
Your answer
Watch out 36 two-digit pages tempts 10 + 36, but page 10 is already the first — so 45.
How to solve
Strategy Work Backwards — The finished total is given and the starting page is known, so this is an end-state problem: I spend the 81 digits from the front instead of counting up page by page. Pages do not all cost the same, so I first split them into blocks that cost the same each — the one-digit pages at 1 digit apiece and the two-digit pages at 2 digits apiece. Paying for the cheap block first tells me exactly how many digits are left for the next block, and dividing gives the number of pages that block buys.
1STEP 1
Split the pages into same-cost blocks
Group the pages into one-digit and two-digit.
Sorting the pages by how many digits their numbers need turns one long count into two short ones, and each short one is just equal groups.
4.OA.A.3Identify SubproblemsSorting the pages by how many digits their numbers need turns one long count into two short counts of equal groups.
Why?
Every page number has one definite digit length, so it falls into exactly one block and never into two.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
Why?
Inside one block every page costs the same number of digits, so that block's cost is a count of equal groups.
🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.
2STEP 2
Pay for the one-digit pages
Pages 1 to 9 use 9 digits.
9 × 1 = 9
Nine groups of one is a multiplication a fourth grader does instantly, and it clears the whole first block in one line.
4.NBT.B.5Identify Subproblems3STEP 3
See how many digits are left
That leaves 81 − 9 = 72.
81 - 9 = 72
Starting from the finished total and taking away what is already accounted for is the same 'undo' move used to find a missing addend.
4.OA.A.3Work Backwards4STEP 4
Turn the leftover digits into a page count
Two-digit pages number 72 ÷ 2 = 36.
72 ÷ 2 = 36
'How many groups of 2 fit in 72' is exactly what division answers, and a zero remainder confirms nothing is left dangling.
4.NBT.B.6Work Backwards5STEP 5
Name the 36th two-digit page
Counting 36 on from page 10 lands on page 45.
10 + 36 - 1 = 45
Listing the first few — 10 is 1st, 11 is 2nd, 12 is 3rd — makes the 'add one less than you think' rule visible instead of something to memorise.
4.OA.A.3Make A Systematic List6STEP 6
Check the digit total for 45 pages
9 + 36 × 2 = 81 checks out.
9 + 36 × 2 = 9 + 72 = 81
Rebuilding the total forwards is an independent check: if the forward count lands on 81 too, the backwards reasoning had no off-by-one slip.
4.NBT.B.5Make A Systematic ListAnswer
45 pages
9 + 36 × 2 = 81
The count is of digits, so 81 digits must buy fewer than 81 pages once numbers start costing 2 digits apiece — and 45 is indeed well under 81. It is also sensible that the book stops in the two-digit range: reaching page 100 would already need 9 + 90 × 2 + 3 = 192 digits, far more than 81, while pages 1 to 9 alone would use only 9. So the last page has to be a two-digit page, and 45 sits comfortably there.
Takeaway
Page numbers are not all the same price in digits, so count them in blocks — the cheap one-digit pages first, then see how many two-digit pages the leftover digits can buy.
- Split the pages into same-cost blocks
- Pay for the one-digit pages
- See how many digits are left
- Turn the leftover digits into a page count
- Name the 36th two-digit page
- Check the digit total for 45 pages