AMC 10 · 2003 · #20

Grade 7 probability
Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region
base-conversionprobability-basicinterval-arithmetic identify-subproblems ↑ Prerequisites: base-conversion
📏 Long solution 💡 3 insights
Problem
Pick a three-digit base-10 number n (that is, from 100 to 999) at random. We want the chance that when n is rewritten in base 9 it still has exactly three digits, and when rewritten in base 11 it also has exactly three digits. Then say which listed value the chance is closest to.

Pick an answer.

(A)
0.3
(B)
0.4
(C)
0.5
(D)
0.6
(E)
0.7

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Each base gives its own separate condition, so Tool #7 (Identify Subproblems) says: figure out the exact range of n that is three-digit in base 9, do the same for base 11, then keep only the n that satisfy both plus the given base-10 condition. To turn 'three digits in base b' into a clean range we use Tool #14 (Extreme Principle): the smallest and largest three-digit numbers in a base are the boundaries, and every base power sits right at those edges. Finally Tool #1 (Draw a Diagram) — a number line of the three overlapping ranges — makes the shared stretch obvious, so we can just count it and divide.

1STEP 1

What 'three digits in base b' means

In base b the smallest three-digit numeral is b² and the largest is one below b³, so three digits means b² ≤ n ≤ b³-1.

three digits in base b ⇔ b² ≤ n ≤ b³-1
2STEP 2

Write down each base's range

Plug the boundaries in: base 9 gives 81 ≤ n ≤ 728, base 11 gives 121 ≤ n ≤ 1330, and the pool is 100 ≤ n ≤ 999.

81 ≤ n ≤ 728 (base 9), 121 ≤ n ≤ 1330 (base 11), 100 ≤ n ≤ 999 (base 10)
3STEP 3

Overlap the three ranges

All three hold at once only from the largest floor to the smallest ceiling — floors give 121, ceilings give 728, so 121 ≤ n ≤ 728.

max(81,121,100)=121, min(728,1330,999)=728 → 121 ≤ n ≤ 728
4STEP 4

Count the winners and divide

The window holds 728-121+1 = 608 of the 900 numbers, so the probability is about 0.676 — closest to 0.7, choice (E).

(728-121+1)/900 = 608/900 ≈ 0.676 → (E) 0.7
Answer
0.7
The favorable window 121 to 728 sits comfortably inside the 100 to 999 pool, and it covers a bit more than two-thirds of that span (608 out of 900), so a probability near 0.68 makes sense. It cannot be tiny or near 1, which rules out the low choices; and since 608/900 rounds up past 0.65, 0.7 beats 0.6. A quick sanity check on the bounds: 120 in base 11 is AA₁₁ (two digits) but 121 = 100₁₁ (three digits), and 729 = 1000₉ (four digits) while 728 = 888₉ (three digits), confirming the edges are exactly 121 and 728.
💡Key takeaway

A number has three digits in base b exactly when it runs from b² up to b³-1, so to satisfy several bases at once you take the highest floor and the lowest ceiling and count what is left.

  • What 'three digits in base b' means
  • Write down each base's range
  • Overlap the three ranges
  • Count the winners and divide

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