AMC 10 · 2015 · #16

Grade 7 probability

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

probability-basicmultiplespermutations-basic systematic-enumeration ↑ Prerequisites: probability-basicmultiples

📏 Medium solution 💡 3 insights

Problem

Al, Bill, and Cal each get a different whole number from 1 to 10, all chosen at random. Find the probability that Al's number is a whole-number multiple of Bill's number and Bill's number is a whole-number multiple of Cal's number.

Pick an answer.

(A)

$\dfrac{9}{1000}$

(B)

$\dfrac{1}{90}$

(C)

$\dfrac{1}{80}$

(D)

$\dfrac{1}{72}$

(E)

$\dfrac{2}{121}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The condition links the three numbers by divisibility, so first reorganize it (Tool #15): since the numbers are distinct and each divides the next, Cal's number is strictly smaller than Bill's, which is strictly smaller than Al's. That turns the question into counting divisor chains. Split the work into two subproblems (Tool #7): how many favorable chains exist, and how many total assignments exist. The favorable chains are few and structured, so a systematic list (Tool #2) catches every one without missing or repeating.

1STEP 1

Turn the rule into an order

Since the numbers differ and each divides the next, a divisor must be smaller — so the chain is forced to Cal ＜ Bill ＜ Al.

Cal ∣ Bill ∣ Al, Cal ＜ Bill ＜ Al

2STEP 2

Count all equally likely assignments

Ignore the rule and deal three distinct numbers: 10 · 9 · 8 = 720 equally likely assignments form the denominator.

10 · 9 · 8 = 720

3STEP 3

List every divisor chain

List chains by Cal: Cal =1 gives 4+2+1+1 = 8, Cal =2 gives 1, Cal ≥ 3 gives none — 9 favorable chains in all.

4+2+1+1_Cal=1 + 1_Cal=2 = 9

4STEP 4

Divide favorable by total

Each chain is exactly one assignment, so 9/720 = 1/80 — choice (C).

9/720 = 1/80 → (C)

Answer

1/80

The favorable count 9 is tiny next to 720, which fits a strict double condition that most random trios fail, so a probability near 1/80 is sensible. Spot-checking the chain (1,2,8): 8 is a multiple of 2 and 2 is a multiple of 1, and all are distinct, so it correctly counts. The reduced fraction 1/80 also sits cleanly among the answer choices.

💡Key takeaway

When distinct numbers must each divide the next, they have to climb in size — so just list the few chains that fit and divide by all the ways to deal the numbers out.

Turn the rule into an order
Count all equally likely assignments
List every divisor chain
Divide favorable by total

Turn the rule into an order

Count all equally likely assignments

List every divisor chain

Divide favorable by total

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