AMC 8 · 2009 · #13
Grade 7 probabilityProblem
A three-digit integer contains one of each of the digits 1, 3, and 5. What is the probability that the integer is divisible by 5?
Pick an answer.
AMC 8 2009 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: A three-digit integer is formed using each of the digits $1$, $3$, and $5$ exactly once. If the arrangement is chosen at random, what is the probability that the resulting number is divisible by $5$?
Givens: The digits used are exactly $\{1, 3, 5\}$, each appearing once; The number has three digits — hundreds, tens, ones; Every distinct three-digit arrangement is equally likely; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{5}{6}$
Unknowns: The probability that the random three-digit number is divisible by $5$
Understand
Restated: A three-digit integer is formed using each of the digits $1$, $3$, and $5$ exactly once. If the arrangement is chosen at random, what is the probability that the resulting number is divisible by $5$?
Givens: The digits used are exactly $\{1, 3, 5\}$, each appearing once; The number has three digits — hundreds, tens, ones; Every distinct three-digit arrangement is equally likely; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{5}{6}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Your Focus
There are only $3! = 6$ arrangements of three digits, so Tool #2 (Make a Systematic List) can write them all out and we directly count the favorable ones — no formula needed. Tool #16 (Change Your Focus) sharpens the question: instead of checking each whole number for divisibility, refocus on the ones digit alone, since a number is divisible by $5$ exactly when its ones digit is $0$ or $5$. With the focus on the last digit, the counting collapses to "how many of the $6$ arrangements end in $5$?"
Execute — Answer: B
3.OA.A.1 Step 1 - List every three-digit number that uses each of $1, 3, 5$ once.
- Fix the hundreds digit, then permute the other two; sweep through hundreds $= 1, 3, 5$ in order so nothing is missed or repeated.
💡 Listing $3$ groups of $2$ ($2$ arrangements per fixed hundreds digit) is the Grade 3 meaning of multiplication: $3 \times 2 = 6$.
4.OA.B.4 Step 2 - Switch focus from the whole number to its ones digit.
- The divisibility-by-$5$ test says a whole number is divisible by $5$ iff its last digit is $0$ or $5$.
- Since $0$ is not one of our digits, the favorable arrangements are exactly the ones whose ones digit is $5$.
💡 The Grade 4 divisibility rule for $5$ depends only on the last digit, so we can ignore the hundreds and tens places.
3.OA.A.1 Step 3 - Mark the arrangements from the list whose ones digit is $5$.
- Scanning $\{135, 153, 315, 351, 513, 531\}$, the numbers ending in $5$ are $135$ and $315$.
💡 With the ones digit pinned to $5$, the hundreds and tens slots are filled by $1$ and $3$ in either order — $2$ ways.
7.SP.C.7 Step 4 Form the probability as favorable over total and simplify by dividing top and bottom by the common factor $2$.
💡 Each of the $6$ arrangements is equally likely, so probability is just the fraction of arrangements that are favorable.
3.OA.A.1 List every three-digit number that uses each of $1, 3, 5$ once. Fix the hundreds 4.OA.B.4 Switch focus from the whole number to its ones digit. The divisibility-by-$5$ te 3.OA.A.1 Mark the arrangements from the list whose ones digit is $5$. Scanning ${135, 15 7.SP.C.7 Form the probability as favorable over total and simplify by dividing top and bo Review
Reasonableness: Quick sanity check by symmetry. Each of the three digits $1, 3, 5$ is equally likely to land in the ones slot, so the chance that any specific digit (here, $5$) is the ones digit is exactly $\tfrac{1}{3}$. That matches the listed count $\tfrac{2}{6} = \tfrac{1}{3}$ and matches choice (B). The other choices are easy to rule out: (A) $\tfrac{1}{6}$ would mean only one arrangement works, (C)–(E) would need at least three arrangements ending in $5$, but only $2$ exist.
Alternative: Tool #5 (Find a Pattern) gives a one-line solution without listing. In a random arrangement of the three distinct digits, every digit has an equal $\tfrac{1}{3}$ chance of occupying the ones place. The number is divisible by $5$ iff the ones digit is $5$, and $P(\text{ones} = 5) = \tfrac{1}{3}$ — answer (B). This generalizes: for $n$ distinct nonzero digits with exactly one of them equal to $5$, the probability of divisibility by $5$ is $\tfrac{1}{n}$.
CCSS standards used (min grade 7)
3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Counting the $6$ arrangements as $3$ choices for the hundreds digit times $2$ arrangements of the remaining two digits, and counting the $2$ favorable arrangements as $1$ choice (ones $= 5$) times $2$ arrangements of $\{1, 3\}$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Applying the divisibility-by-$5$ rule (last digit is $0$ or $5$) to decide which arrangements count as favorable.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating each of the $6$ arrangements as an equally likely outcome and computing $P(\text{divisible by }5) = \tfrac{2}{6} = \tfrac{1}{3}$.)
⭐ This AMC 8 problem only needs Grade 7 probability — list every arrangement, use the Grade 4 divisibility rule for $5$, then divide favorable by total.
⭐ This AMC 8 problem only needs Grade 7 probability — list every arrangement, use the Grade 4 divisibility rule for $5$, then divide favorable by total.
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