Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #24
Grade 7 probabilityPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 2/4 = 1/2.
Reduce to the digit sum
Divisibility by 3 depends only on the digit sum, so reordering the same three digits never changes the answer.
The Grade 4 multiples-and-factors rule says: to test divisibility by 3, add the digits. Since reshuffling the same three digits keeps the sum, every arrangement of a winning subset is also a winner.
Whether the three-digit number is a multiple of 3 depends only on which three digits are drawn, not on the order they come out.
▸ Why?
The number is a multiple of 3 exactly when its three digits add to a multiple of 3, and reshuffling the same digits never changes that sum.
▸ Why?
A three-digit number leaves the same remainder as its digit sum when divided by 3, so one is a multiple of 3 exactly when the other is.
▸ Why?
By place value the number is (hundreds digit) times 100 plus (tens digit) times 10 plus (ones digit); since 100 and 10 are each one more than a whole batch of 3, peeling those batches off leaves exactly the three digits behind.
▸ Why?
Once the whole batches of 3 are set aside, what is left over is just the digit sum, so the number and its digit sum share the same remainder on division by 3 — and the number is a multiple of 3 precisely when that remainder is 0.
▸ Why?
Reordering the digits just adds the same three values in a different order, and a sum is the same in any order.
Count subsets instead of orders
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.
Each three-digit subset arises from the same number of ordered draws (3! = 6), so the 24 ordered outcomes split evenly into 4 groups of 6 — and within a group, all six orderings agree on divisibility by 3.
7.SP.C.7Change Focus Count The ComplementCheck each subset's sum
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.
Going in order (smallest missing digit: 4, 3, 2, 1) catches every subset exactly once. Two of the four sums — 6 and 9 — are multiples of 3.
7.SP.C.8Make A Systematic ListForm the probability
Two of four equally likely subsets win, so the probability is = , choice (C).
Equally likely outcomes mean probability equals the favorable fraction. Half the subsets win, so the answer is one-half.
7.SP.C.7Change Focus Count The ComplementDivisibility 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 .
- Reduce to the digit sum
- Count subsets instead of orders
- Check each subset's sum
- Form the probability
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