AMC 8 · 2009 · #13

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

probability-basicpermutations-basicdivisibility-rules systematic-enumeration ↑ Prerequisites: multi-digit-arithmeticdivisibility-rules

📏 Short solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{1}{3}$

(C)

$\frac{1}{2}$

(D)

$\frac{2}{3}$

(E)

$\frac{5}{6}$

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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?"

1STEP 1

List all arrangements of 1, 3, 5: fixing each hundreds digit gives 6 three-digit numbers in all.

{135, 153, 315, 351, 513, 531} — total = 6

2STEP 2

Switch focus to the ones digit: divisibility by 5 means ending in 0 or 5, and with no 0 here it means the ones digit is 5.

divisible by 5 ⇔ ones digit ∈ {0, 5} → ones digit = 5

3STEP 3

Scan the list for numbers ending in 5: only 135 and 315 qualify, giving 2 favorable arrangements.

favorable = {135, 315} → count = 2

4STEP 4

Divide favorable by total and simplify $\frac{2}{6}$ to get $\frac{1}{3}$ , which is choice (B).

P =

\frac{2}{6}

\frac{1}{3}

→ (B)

Answer

\frac{1}{3}

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

\frac{1}{3}

. That matches the listed count

\frac{2}{6}

\frac{1}{3}

and matches choice (B). The other choices are easy to rule out: (A)

\frac{1}{6}

would mean only one arrangement works, (C)–(E) would need at least three arrangements ending in 5, but only 2 exist.

💡Key takeaway

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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