AMC 8 · 2009 · #22

Grade 5 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

systematic-enumerationdigit-constraintsplace-valuecomplementary-counting complementary-countingcasework ↑ Prerequisites: multi-digit-arithmeticplace-value

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

Count the whole numbers strictly between 1 and 1000 whose decimal representation contains no digit "1". So we need to count integers n with 1 ＜ n ＜ 1000 such that none of the digits of n is 1.

Pick an answer.

(A)

512

(B)

648

(C)

720

(D)

728

(E)

800

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Count the Possibilities

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.

1STEP 1

Pad every number 0–999 to a 3-digit string, so each of the 3 positions independently allows 9 digits (all but 1).

allowed digits per slot = 10 - 1 = 9

2STEP 2

Multiply the three independent 9-way choices, giving 729 padded strings with no digit 1.

9 × 9 × 9 = 729

3STEP 3

Subtract the lone string 000 (it means 0, out of range) — 1 excludes itself and 1000 has four digits — leaving 728.

729 - 1 = 728

4STEP 4

The count 728 matches answer choice (D).

728 → (D)

Answer

728

Roughly

\frac{9}{10}

of digits are allowed in each of 3 slots, so the fraction of 3-digit strings that avoid 1 is about (

\frac{9}{10}

)³ = 0.729. Out of the 1000 strings from 000 to 999, that predicts about 729 — matching our 728 exactly after subtracting 000. The other choices fail this sanity check: 512 = 8³ would forbid two digits, 800 would forbid only fractions of slots, etc.

💡Key takeaway

Pad 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.

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