AMC 10 · 2017 · #16
Easy mode Grade 4Look at every whole number from 1 to 2017. How many of them have at least one 0 somewhere in their digits?
Pick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count how many whole numbers from 1 up to 2017 have at least one digit that is 0 when written normally in base ten.
Givens: We look at every positive integer from 1 to 2017; A number counts if its ordinary base-ten spelling shows the digit 0 somewhere
Unknowns: How many of these 2017 numbers contain a 0 digit
Understand
Restated: Count how many whole numbers from 1 up to 2017 have at least one digit that is 0 when written normally in base ten.
Givens: We look at every positive integer from 1 to 2017; A number counts if its ordinary base-ten spelling shows the digit 0 somewhere
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Counting numbers that have a 0 directly forces messy 'at least one' casework. Flip it: count the numbers with NO zero at all, which is clean because each digit just has to avoid 0. Split that count by how many digits the number has, count each group with a simple multiply rule, then subtract from the total 2017.
Execute — Answer: A
4.OA.A.3 Step 1 Total count and the flip
- There are exactly 2017 numbers to check.
- Instead of counting the ones that contain a 0, count the ones that contain no 0 at all.
- Every number is in exactly one of these two groups, so (contains a 0) = 2017 - (contains no 0).
- Group the no-zero count by digit length: 1-digit, 2-digit, 3-digit, and the 4-digit numbers up to 2017.
💡 Counting the opposite group is easier here, and the total minus the opposite gives what you want.
4.NBT.A.2 Step 2 One-digit numbers
- The one-digit numbers are 1 through 9.
- None of them is 0 and none contains a 0 digit, so all 9 of them have no zero.
💡 Single digits 1 to 9 simply have no place where a 0 could hide.
3.OA.A.1 Step 3 Two- and three-digit numbers with no zero
- For a two-digit number, the tens digit can be 1-9 (9 choices) and the units digit must also avoid 0, so 1-9 (9 choices): 9 x 9 = 81 zero-free numbers.
- For a three-digit number, each of the three places independently picks from 1-9: 9 x 9 x 9 = 729 zero-free numbers.
💡 Independent choices multiply, so 9 options per digit gives 9 to the power of the number of digits.
4.NBT.A.2 Step 4 Four-digit numbers up to 2017 with no zero
- The four-digit numbers run 1000 to 2017.
- Those starting with 1 (1000-1999) are zero-free when the last three digits each avoid 0: 9 x 9 x 9 = 729, and these all sit below 2000.
- Those starting with 2 (2000-2017) all have a 0 in the hundreds place, so none of them is zero-free.
- So the four-digit range contributes 729 zero-free numbers.
💡 Every number from 2000 to 2017 already shows a 0 right after the 2, so it can never be zero-free.
4.NBT.B.4 Step 5 Add the zero-free counts and subtract
- Total zero-free numbers from 1 to 2017: 9 + 81 + 729 + 729 = 1548.
- The numbers that DO contain a 0 are everything else: 2017 - 1548 = 469.
- That matches choice (A).
💡 Take the whole pile and remove the zero-free ones; what is left is exactly the numbers that hide a 0.
4.OA.A.3 There are exactly 2017 numbers to check. Instead of counting the ones that conta 4.NBT.A.2 The one-digit numbers are 1 through 9. None of them is 0 and none contains a 0 d 3.OA.A.1 For a two-digit number, the tens digit can be 1-9 (9 choices) and the units digi 4.NBT.A.2 The four-digit numbers run 1000 to 2017. Those starting with 1 (1000-1999) are z 4.NBT.B.4 Total zero-free numbers from 1 to 2017: 9 + 81 + 729 + 729 = 1548. The numbers t Review
Reasonableness: About a quarter of the numbers up to 2017 having a 0 feels right: 469 out of 2017 is roughly 23%. A rough estimate agrees, since for three-digit numbers the chance of being zero-free is (9/10)(9/10) = 81% per the last two digits, leaving about 19% with a zero, and the four-digit band 2000-2017 is entirely zero-containing, nudging the share up. The five answer choices cluster near 470, and only 469 comes out of an exact zero-free count of 1548.
Alternative: Count the zero-containing numbers directly by place. Among two-digit numbers, 9 end in 0 (10,20,...,90). Among three-digit numbers, 9x9=81 have a 0 in some single spot plus more with two zeros - this inclusion-exclusion is doable but error-prone, which is exactly why the complement (count the zero-free ones and subtract) is the cleaner route to 469.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Reframing the count as total minus the zero-free count)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Reasoning about each digit place and the 2000-2017 range)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Multiplying 9 choices per digit to count zero-free numbers)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the group counts and subtracting from 2017)
⭐ When 'at least one' is hard to count, count the opposite (none at all) and subtract from the total.
⭐ When 'at least one' is hard to count, count the opposite (none at all) and subtract from the total.
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