Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #9
Grade 4 rate-rationumber-theoryPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Instead of writing an equation with an unknown total, we use Tool #9: replace the abstract "total" with the smallest concrete totals that work, and check each one. Because 1/3 and 1/4 of the total must both be whole numbers, the total has to be a multiple of 3 and of 4 — so the candidates are 12, 24, 36, …. Tool #6 (Guess and Check) walks through these smallest totals in order; the first total that leaves a non-negative number of yellow marbles is the answer. Tool #3 (Eliminate) keeps us honest: once we find Y = 4, we can confirm that the smaller answer choices (1, 2, 3) are impossible.
Decide which totals work
Blue needs and red needs to be whole, so the total is a multiple of both 3 and 4 — candidates are 12, 24, 36, ...
Listing common multiples of 3 and 4 is exactly the Grade 4 "multiples" skill — no algebra needed.
The only totals that can work are the numbers that both 3 and 4 divide evenly — 12, 24, 36, and so on.
▸ Why?
Exactly one-third of the marbles are blue, and a count of marbles is a whole number, so the whole pile must split into three equal whole groups — meaning 3 divides the total.
▸ Why?
Taking one-third of the total gives the blue count, and the total is just that whole blue count repeated three times — three equal groups.
▸ Why?
Exactly one-fourth of the marbles are red, and a count of marbles is a whole number, so the whole pile must split into four equal whole groups — meaning 4 divides the total.
▸ Why?
Taking one-fourth of the total gives the red count, and the total is just that whole red count repeated four times — four equal groups.
▸ Why?
The total has to break into three equal groups and into four equal groups at the same time, so it must be a number that both 3 and 4 divide — a common multiple of 3 and 4.
▸ Why?
The smallest number that both 3 and 4 divide evenly is 12, and the numbers that both divide are then exactly the multiples of that 12 — 12, 24, 36, and onward — because two step-counting cycles first line up at their least common multiple and keep lining up only at its multiples.
Test a total of 12
Test total 12: blue 4 + red 3 + green 6 = 13, already 1 over 12 before any yellow — so 12 is impossible.
Multiplying and adding small whole numbers is Grade 3 — perfect for plugging in a guess and checking the fit.
3.OA.A.3Guess And CheckTest a total of 24
Test total 24: blue 8 + red 6 + green 6 = 20, leaving 24 - 20 = 4 yellow — a valid non-negative whole number.
Multi-step word problems with the four operations within 100 are Grade 3 — no fractions to add, just careful arithmetic.
3.OA.D.8Guess And CheckConfirm 4 is smallest
The only smaller total, 12, already failed and bigger totals like 36 give 9 yellow, so 4 is the smallest.
Sweeping through valid totals in order and stopping at the first success is the Grade 4 "multi-step word problem with reasoning about results" move.
4.OA.A.3Eliminate PossibilitiesMatch the answer choice
The smallest number of yellow marbles is 4, matching choice (D).
Reading off the answer choice that matches 4 uses only Grade 2 number recognition.
2.NBT.B.5Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 thinking about multiples of 3 and 4 that you already know!
- Decide which totals work
- Test a total of 12
- Test a total of 24
- Confirm 4 is smallest
- Match the answer choice
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