AMC 10 · 2003 · #20
Grade 7 probabilityA 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?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $n$ is a randomly chosen three-digit base-10 number, so $100 \le n \le 999$; We need the base-9 form of $n$ to have exactly three digits; We need the base-11 form of $n$ to have exactly three digits; Answer choices: (A) $0.3$, (B) $0.4$, (C) $0.5$, (D) $0.6$, (E) $0.7$
Unknowns: The set of $n$ that are three-digit in base $9$ AND base $11$ at the same time; The probability (favorable count) $\div$ (total count), rounded to the nearest listed value
Understand
Restated: 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.
Givens: $n$ is a randomly chosen three-digit base-10 number, so $100 \le n \le 999$; We need the base-9 form of $n$ to have exactly three digits; We need the base-11 form of $n$ to have exactly three digits; Answer choices: (A) $0.3$, (B) $0.4$, (C) $0.5$, (D) $0.6$, (E) $0.7$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #14 Extreme Principle, #1 Draw a Diagram
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.
Execute — Answer: E
6.EE.A.1 Step 1 What 'three digits in base b' means
- In base $b$, the smallest three-digit numeral is $100_b$, which equals $b^2$.
- The largest is the numeral with three copies of the top digit $b-1$, and that value is one less than $b^3$.
- So a number has exactly three digits in base $b$ precisely when it sits from $b^2$ up to $b^3-1$.
- This single fact is the key that unlocks every base in the problem.
💡 The first three-digit number is where the $b^2$ place first turns on, and the last is just before the $b^3$ place turns on.
6.EE.B.8 Step 2 Write down each base's range
- Plug the boundaries in.
- Base $9$: $9^2=81$ and $9^3-1=728$, so three-digit means $81 \le n \le 728$.
- Base $11$: $11^2=121$ and $11^3-1=1330$, so three-digit means $121 \le n \le 1330$.
- Base $10$ is the given pool: $100 \le n \le 999$.
- We now have three plain ranges instead of three word conditions.
💡 Each 'three-digit' rule is just a low end and a high end, so replace the words with an interval.
6.NS.C.7 Step 3 Overlap the three ranges
- For all three to hold at once, $n$ must be at least the largest of the three low ends and at most the smallest of the three high ends.
- The low ends are $81$, $121$, $100$, and the largest is $121$.
- The high ends are $728$, $1330$, $999$, and the smallest is $728$.
- So the base-11 floor and the base-9 ceiling are the ones that bind, and every valid $n$ lies from $121$ to $728$.
- Picture three bars on a number line: the stretch they all cover runs from $121$ to $728$.
💡 An 'all three at once' window starts at the highest starting line and ends at the lowest finish line.
7.SP.C.5 Step 4 Count the winners and divide
- The whole numbers from $121$ to $728$ number $728-121+1 = 608$.
- The total pool of three-digit base-10 numbers is $999-100+1 = 900$.
- So the probability is $\tfrac{608}{900} \approx 0.676$.
- Among the choices, $0.676$ is closest to $0.7$, so the answer is (E).
💡 Probability is just how many numbers win out of how many are in the hat.
6.EE.A.1 In base $b$, the smallest three-digit numeral is $100_b$, which equals $b^2$. Th 6.EE.B.8 Plug the boundaries in. Base $9$: $9^2=81$ and $9^3-1=728$, so three-digit means 6.NS.C.7 For all three to hold at once, $n$ must be at least the largest of the three low 7.SP.C.5 The whole numbers from $121$ to $728$ number $728-121+1 = 608$. The total pool o Review
Reasonableness: 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 $AA_{11}$ (two digits) but $121 = 100_{11}$ (three digits), and $729 = 1000_9$ (four digits) while $728 = 888_9$ (three digits), confirming the edges are exactly $121$ and $728$.
Alternative: Skip naming all three ranges and just find the binding pair directly. The base with the largest square gives the tightest floor, so base $11$ sets the low edge at $11^2 = 121$; the base with the smallest cube gives the tightest ceiling, so base $9$ sets the high edge at $9^3 - 1 = 728$ (base 10's own limits $100$ and $999$ are looser than both). That immediately gives the window $121 \le n \le 728$ and the same count $608$ over $900 \approx 0.68$.
CCSS standards used (min grade 7)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating $b^2$ and $b^3-1$ to turn 'three digits in base $b$' into number boundaries.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Writing each base's three-digit condition as an interval like $81 \le n \le 728$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Comparing the low ends and high ends to find the tightest floor ($121$) and ceiling ($728$) of the shared window.)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Forming the probability $608/900 \approx 0.676$ as favorable count over total count.)
⭐ A number has three digits in base $b$ exactly when it runs from $b^2$ up to $b^3-1$, so to satisfy several bases at once you take the highest floor and the lowest ceiling and count what is left.
⭐ A number has three digits in base $b$ exactly when it runs from $b^2$ up to $b^3-1$, so to satisfy several bases at once you take the highest floor and the lowest ceiling and count what is left.
More like this
Same archetype — closest grade level first.