Problem
Count how often each digit appears
Write out every whole number from 1 to 99. Fill in how many times each digit 0 to 9 gets written. Count writings, not how many numbers contain the digit. Give the count for all ten digits.
Reasoning · Grade 4-1 Numbers and Digit Counts
Your answer
Watch out All ten digits look like 20 each, but 0 cannot lead and appears just 9 times.
How to solve
Strategy Make a Systematic List — Writing out all 99 numbers and tallying by hand would take forever and invite mistakes, so instead of sorting the digits by which number they came from, I re-sort them by which place they were written in. Every digit written on the page sits either in a ones place or in a tens place, and inside each place the digits march in a plain repeating order that is easy to list. Counting each place separately and then adding the two counts keeps the job small and systematic.
1STEP 1
See how many digits get written altogether
Altogether 189 digits get written.
9 × 1 + 90 × 2 = 9 + 180 = 189
Splitting the list into the one-digit numbers and the two-digit numbers is just counting how many boxes there are to fill, and counting from 10 to 99 gives 90 numbers the same way counting from 1 to 90 does.
3.NBT.A.2Make A Systematic List2STEP 2
Sort the written digits by place instead of by number
Regroup them by place instead of by number.
47 → tens digit 4, ones digit 7
Reading a two-digit number as tens-and-ones is exactly what a young solver already does when reading 47 as forty-seven, and it turns one messy pile of digits into two tidy ones.
1.NBT.B.2Organize Information In More WaysSorting the written digits by place rather than by number turns one messy pile into two tidy ones.
Why?
A two-digit number is a count of tens together with a count of ones, so each written digit belongs to one place and one only.
🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.
Why?
The two place-lists cover every written digit and never overlap, so their counts can simply be added at the end.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
3STEP 3
Count the ones digits
Each digit shows up 10 times in the ones place.
3, 13, 23, 33, 43, 53, 63, 73, 83, 93 → 10 threes
The ones digits cycle 0 through 9 forever, so once a solver sees one full lap they can count laps instead of counting numbers.
4.OA.C.5Look For A Pattern4STEP 4
Count the tens digits
In the tens place 1 to 9 each show up 10 times.
30, 31, 32, 33, 34, 35, 36, 37, 38, 39 → 10 threes in the tens place
Nine blocks of ten numbers, one block per leading digit, is the same 'equal groups' idea used for 9 x 10, so no new skill is needed.
3.OA.A.1Make A Systematic List5STEP 5
Add the two places together and fill the table
Adding gives 20 each for 1 to 9, and 9 for 0.
10 + 10 = 20 (digits 1–9), 9 + 0 = 9 (digit 0)
Because the two place-lists never overlap, the totals can simply be added — the same reason a solver can count red cubes and blue cubes separately and add.
3.NBT.A.2Organize Information In More Ways6STEP 6
Check against the total from step 1
The totals add back to 189.
9 + 9 × 20 = 9 + 180 = 189
Every digit written must be counted exactly once, so the ten counts have to rebuild the total of 189 — a check a solver can do without redoing any of the work.
3.NBT.A.2Make A Systematic ListAnswer
0 nine times, each of 1 to 9 twenty times
9 + 9 × 20 = 189
The ten counts add to 9 + 9 x 20 = 189, which matches the digit total found independently from 9 one-digit numbers and 90 two-digit numbers, so nothing was counted twice or missed. It is also sensible that 0 is the odd one out and is smaller than the rest: 0 can never lead a number, so it loses the whole tens place, and its ones-place list starts at 10 rather than at 0. Every count is a whole number well under 189, as it must be.
Takeaway
Count the ones column and the tens column separately — then every digit from 1 to 9 shows up 20 times, and only lonely 0 gets left out of the tens!
- See how many digits get written altogether
- Sort the written digits by place instead of by number
- Count the ones digits
- Count the tens digits
- Add the two places together and fill the table
- Check against the total from step 1