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2003 AMC 10A Problem 20

2003 AMC 10A Problem 20AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 3 of 5 · about 3 min · position P16-20

A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?

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

Answer

Key insight

Three digits in base b means b^2 <= n < b^3, so n must lie in [121, 728]; that is 608 of the 900 three-digit numbers, about 0.68.

Solution

A positive integer has exactly three digits in base $b$ when $b^2 \le n \le b^3 - 1$ .

Base $9$ : $81 \le n \le 728$ .
Base $11$ : $121 \le n \le 1330$ .
Base $10$ (the sample space): $100 \le n \le 999$ , which is $900$ numbers.

All three conditions hold exactly when $121 \le n \le 728$ , an interval of $728 - 121 + 1 = 608$ integers.

The probability is $\dfrac{608}{900} \approx 0.676$ , closest to $0.7$ .

The answer is $\boxed{\textbf{(E)}\ 0.7}$ .

Why this works

Digit counts are interval conditions: $k$ digits in base $b$ is the range $[b^{k-1}, b^k - 1]$ . Intersecting the three intervals reduces the problem to counting integers in $[121, 728]$ . The word "closest" signals an estimate is enough: $608/900$ is a bit over two thirds.

The trap

Treating 729 = 9^3 as a three-digit base-9 number (it is 1000 in base 9), or taking 100 rather than 11^2 = 121 as the smallest three-digit base-11 number.

Common mistakes

Treating 729 = 9^3 as a three-digit base-9 number (it is 1000 in base 9), or taking 100 rather than 11^2 = 121 as the smallest three-digit base-11 number.
Rounding $0.676$ down to $0.6$ or dividing by $999$ instead of $900$ .

Techniques

Bound the quantity above/below or estimate to pin it down