Competition · AMC preparation · step 4 of 4
AMC 8 · 2009 · #22
Grade 5 countingPick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are counting how many length-3 digit strings avoid one specific digit, which is exactly the multiplication principle situation Tool #13 (Count the Possibilities) handles. The clean trick is to pad every number up to three digits (write 7 as "007", 42 as "042", etc.) — then every integer from 0 to 999 becomes a 3-slot string, each slot independently picks from {0,1,2,…,9}, and we just forbid digit 1 in every slot. Tool #7 (Identify Subproblems) handles the small endpoint cleanup: the padded count includes 000 (which is not in our range) but excludes nothing else we want, so a single subtraction fixes it.
Pad every number to 3 digits
Pad every number 0–999 to a 3-digit string, so each of the 3 positions independently allows 9 digits (all but 1).
"9 choices, repeated for each independent slot" is the same equal-groups multiplication idea taught in Grade 3.
3.OA.A.1Convert To AlgebraMultiply the slot choices
Multiply the three independent 9-way choices, giving 729 padded strings with no digit 1.
Three independent 9-way choices multiply, just like the Grade 5 standard multi-digit multiplication fluency.
The number of three-slot digit strings that use no 1 is 9 × 9 × 9.
▸ Why?
Each of the three slots may hold any of the ten digits 0–9 except the single banned 1; the ten digits divide with no gap and no overlap into that one forbidden digit and the rest, so exactly 10 - 1 = 9 stay allowed in every slot.
▸ Why?
The three slots are filled independently — the digit chosen in one slot never removes an option from another — so the 9 ways to fill each slot combine by multiplying: each of the 9 first-slot choices pairs with each of the 9 second-slot choices and each of the 9 third-slot choices, giving 9 × 9 × 9.
Adjust for the endpoints
Subtract the lone string 000 (it means 0, out of range) — 1 excludes itself and 1000 has four digits — leaving 728.
Counting then peeling off the one bad case at the boundary is the Tool #7 subproblems pattern, exactly the multi-step word-problem reasoning of Grade 4.
4.OA.A.3Identify SubproblemsMatch against the choices
The count 728 matches answer choice (D).
Choosing the matching option closes the problem.
4.OA.A.3Convert To AlgebraPad every number to three digits, then each slot is a separate 9-choice pick — that is exactly the equal-groups multiplication you learned in Grade 5, with one small fix at the end.
- Pad every number to 3 digits
- Multiply the slot choices
- Adjust for the endpoints
- Match against the choices
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