AMC 10 · 2006 · #21

Grade 4 arithmetic
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
complementary-countingpermutations-basicplace-value identify-subproblems ↑ Prerequisites: complementary-counting
📏 Medium solution 💡 2 insights
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Problem
The four-digit positive integers run from 1000 to 9999. Call one a hit when the digit 2 or the digit 3 shows up in at least one of its four places. How many four-digit integers are hits?

Pick an answer.

(A)
2439
(B)
4096
(C)
4903
(D)
4904
(E)
5416

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The phrase "at least one" is the classic signal for Tool #16 (Count the Complement): counting numbers that contain a 2 or 3 directly forces messy overlapping cases, but the opposite group — numbers with NO 2 and NO 3 — is a single clean count. So the plan is total minus none. Tool #7 (Identify Subproblems) handles each of those two counts by picking one digit position at a time and multiplying the choices. Tool #3 (Eliminate Possibilities) gives a fast sanity check: more than half of all four-digit numbers should qualify, which already points to the largest choices.

1STEP 1

Flip to the complement

Counting "has a 2 or 3" head-on tangles overlaps. Flip it: count every four-digit number, then subtract the ones dodging both digits.

#(at least one 2 or 3) = #(all) - #(no 2 or 3)
2STEP 2

Count all four-digit numbers

Build it one place at a time: 9 digits fit the leading place (not 0) and 10 fit each of the other three, so there are 9000 in all.

9 × 10 × 10 × 10 = 9000
3STEP 3

Count numbers with no 2 and no 3

Ban 2 and 3 everywhere: the leading place keeps 7 digits (no 0, 2, 3) and each other place keeps 8, giving 3584 dodgers.

7 × 8 × 8 × 8 = 7 × 512 = 3584
4STEP 4

Subtract to finish

Take the dodgers away from the whole: 9000 minus 3584 leaves 5416 numbers, each holding a 2 or a 3 — choice (E).

9000 - 3584 = 5416 → (E)
Answer
5416
The answer 5416 is well over half of the 9000 total, which fits intuition: with four digit slots each having a fair chance of landing on a 2 or 3, it is likely that at least one does — so a majority qualifying is expected. This immediately kills (A) 2439 (less than half) and makes the top choices credible. The avoider count checks out too: 8³ = 512 and 7 × 512 = 3584, and 3584 + 5416 = 9000 restores the full total, so no numbers were lost or double-counted.
💡Key takeaway

When a problem says "at least one," count everything and subtract the cases that have none — the leftover is your answer.

  • Flip to the complement
  • Count all four-digit numbers
  • Count numbers with no 2 and no 3
  • Subtract to finish

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