AMC 8 · 2007 · #24

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

probability-basicdigit-sumdivisibility-rulescombinations-basic caseworksystematic-enumeration ↑ Prerequisites: divisibility-rulesprobability-basic

📏 Short solution 💡 2 insights

Problem

A bag holds four slips labeled 1, 2, 3, 4. Draw three slips one at a time without replacement and read them in the order drawn as a three-digit number. What is the probability that this three-digit number is a multiple of 3?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{1}{3}$

(C)

$\frac{1}{2}$

(D)

$\frac{2}{3}$

(E)

$\frac{3}{4}$

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Right Thing

Counting all 4 × 3 × 2 = 24 ordered draws is doable but wasteful, because the digit-sum rule says divisibility by 3 depends only on which digits are used, not on their order. Tool #16 (Change Focus) shifts the count from 24 ordered numbers to the 4 three-element subsets of {1,2,3,4} — each subset is equally likely because each contributes the same 6 orderings. Tool #2 (Systematic List) then enumerates those 4 subsets and checks each digit sum against the divisibility rule. Two subsets work out of four, so the probability is $\frac{2}{4}$ = $\frac{1}{2}$ .

1STEP 1

Divisibility by 3 depends only on the digit sum, so reordering the same three digits never changes the answer.

abc divisible by 3 ⇔ a+b+c divisible by 3

2STEP 2

Instead of all 24 ordered numbers, count the 4 equally likely three-digit subsets of {1,2,3,4}; the probability is (good subsets)/4.

P(multiple of 3) = (good subsets)/4

3STEP 3

Add each subset's digits: {1,2,3}=6 and {2,3,4}=9 are multiples of 3, but {1,2,4}=7 and {1,3,4}=8 are not, so two subsets are good.

{1,2,3}: 6 ✓ {1,2,4}: 7 × {1,3,4}: 8 × {2,3,4}: 9 ✓

4STEP 4

Two of four equally likely subsets win, so the probability is $\frac{2}{4}$ = $\frac{1}{2}$ , choice (C).

P =

\frac{2}{4}

\frac{1}{2}

→ (C)

Answer

\frac{1}{2}

Cross-check by counting ordered numbers directly. Total three-digit numbers: 4 × 3 × 2 = 24. Good subsets are {1,2,3} and {2,3,4}; each contributes 3! = 6 arrangements, for 6 + 6 = 12 winning numbers. Probability =

\frac{12}{24}

\frac{1}{2}

, matching (C). The two counts agree, which confirms that switching to subsets did not change the answer. The result

\frac{1}{2}

is also reasonable: among the four three-digit sums {6, 7, 8, 9}, exactly two are multiples of 3, in line with the rough

\frac{1}{3}

density of multiples — boosted here because the digits cluster near 3 and its multiples.

💡Key takeaway

Divisibility by 3 depends on the digit sum, not the order — so count the four three-digit subsets of {1,2,3,4} instead of all 24 arrangements. Two of the four sums (6 and 9) work, giving probability $\frac{1}{2}$ .

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