AMC 10 · 2017 · #16
Grade 4 arithmeticPick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Total count and the flip
All 2017 numbers split into two groups, so has a 0 = 2017 - has no 0; count the zero-free ones by digit length.
Counting the opposite group is easier here, and the total minus the opposite gives what you want.
4.OA.A.3Count The ComplementOne-digit numbers
The one-digit numbers 1 through 9 have nowhere to hide a 0, so all 9 are zero-free.
Single digits 1 to 9 simply have no place where a 0 could hide.
4.NBT.A.2Identify SubproblemsTwo- and three-digit numbers with no zero
Each digit avoids 0 by picking 1-9, giving 9 x 9 = 81 two-digit and 9 x 9 x 9 = 729 three-digit zero-free numbers.
Independent choices multiply, so 9 options per digit gives 9 to the power of the number of digits.
3.OA.A.1Make A Systematic ListFour-digit numbers up to 2017 with no zero
From 1000-1999 the last three digits each avoid 0 for 9 x 9 x 9 = 729; every number 2000-2017 has a 0 after the 2, adding none.
Every number from 2000 to 2017 already shows a 0 right after the 2, so it can never be zero-free.
4.NBT.A.2Identify SubproblemsAdd the zero-free counts and subtract
Zero-free total 9 + 81 + 729 + 729 = 1548, so the numbers with a 0 are 2017 - 1548 = 469.
Take the whole pile and remove the zero-free ones; what is left is exactly the numbers that hide a 0.
4.NBT.B.4Identify SubproblemsWhen 'at least one' is hard to count, count the opposite (none at all) and subtract from the total.
- Total count and the flip
- One-digit numbers
- Two- and three-digit numbers with no zero
- Four-digit numbers up to 2017 with no zero
- Add the zero-free counts and subtract