Competition · AMC preparation · step 4 of 4
AMC 8 · 2024 · #5
Grade 4 number-theoryPick an answer.
AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only 5 candidate sums and at most a handful of dice pairs for each, so we can simply LIST every pair (d₁, d₂) with 1 ≤ d₁ ≤ d₂ ≤ 6 that adds to a given sum (Tool #2). For each sum we ELIMINATE the candidate the moment we find one pair whose product is a multiple of 6 — that proves the sum is reachable (Tool #3). The whole task splits into the SUBPROBLEM 'is d₁ d₂ divisible by 6?', which we break further into 'is it divisible by 2?' and 'is it divisible by 3?' (Tool #7). No algebra is needed.
Turn 'multiple of 6' into two tests
Since 6 = 2 × 3, a product is a multiple of 6 only when the two dice supply at least one even value and at least one multiple of 3.
Grade 4 students learn that 'multiple of 6' means a number you can build with 2s and 3s, so we split the divisibility test into two easy checks.
4.OA.B.4Identify SubproblemsTest the sum 5
Pairs adding to 5: (1,4)→4 and (2,3)→6. Since 2 × 3 = 6 is a multiple of 6, sum 5 works, so eliminate (A).
Listing every pair that adds to a single-digit number uses Grade 2 fact fluency within 20.
2.OA.B.2Make A Systematic ListTest the sum 6
Pairs adding to 6: (1,5)→5, (2,4)→8, (3,3)→9. None is a multiple of 6, so sum 6 fails every pair.
Multiplying single-digit factors and checking divisibility uses Grade 3 multiplication fluency within 100.
No pair of dice values that adds to 6 can have a product that is a multiple of 6.
▸ Why?
A product is a multiple of 6 only if it holds a factor of 2 and a factor of 3 at the same time, since 6 itself is 2 times 3; so a winning pair must contain both an even value and a multiple of 3.
▸ Why?
A multiple of 6 is a whole number of 6s, and each 6 can be regrouped as a 2 and a 3, so the product always carries a factor of 2 and a factor of 3.
▸ Why?
The only dice values that supply a factor of 3 are 3 and 6, so a pair adding to 6 would have to include one of them, yet each choice fails.
▸ Why?
Counting multiples of 3 by adding 3 over and over gives 3, 6, 9, and up, and only 3 and 6 land inside a die's range of 1 to 6.
▸ Why?
Using a 6 forces its partner to be 6 minus 6, which is 0 and not a value on any die, so the 6 case is impossible.
▸ Why?
Using a 3 forces its partner to be 6 minus 3, which is again 3, so the only remaining pair is 3 and 3, and that pair has no even value to give the needed factor of 2.
▸ Why?
The partner is found by removing the known 3 from the sum 6, and 6 minus 3 is 3 — subtraction just undoes the addition that formed the sum.
▸ Why?
Both dice show 3, which is odd, and an odd number times an odd number stays odd, so the product 9 holds no factor of 2 to supply the even part the pair still needs.
Test the sum 7
Pairs adding to 7: (1,6)→6, (2,5)→10, (3,4)→12. Both 6 and 12 are multiples of 6, so sum 7 works — eliminate (C).
Once you know 6, 12, 18, … are multiples of 6, the check is instant Grade 3 multiplication.
3.OA.C.7Make A Systematic ListTest the sum 8
Pairs adding to 8: (2,6)→12, (3,5)→15, (4,4)→16. Here 2 × 6 = 12 is a multiple of 6, so sum 8 works — eliminate (D).
Listing pairs and multiplying is still Grade 3 work — no algebra needed.
3.OA.C.7Make A Systematic ListTest the sum 9
Pairs adding to 9: (3,6)→18, (4,5)→20. Since 3 × 6 = 18 is a multiple of 6, sum 9 works — eliminate (E).
3 × 6 = 18 = 6 × 3 uses the same Grade 3 multiplication facts.
3.OA.C.7Make A Systematic ListKeep the only sum left
A, C, D, E each yielded an explicit multiple-of-6 pair; only (B) survives, so 6 cannot be the sum.
Putting the casework together to pick the one remaining answer is Grade 3 multi-step problem solving.
3.OA.D.8Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 multiples-and-factors thinking you already know!
- Turn 'multiple of 6' into two tests
- Test the sum 5
- Test the sum 6
- Test the sum 7
- Test the sum 8
- Test the sum 9
- Keep the only sum left
A parent dashboard for the family lives at sensimlab.com.