Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #23
Grade 6 number-theoryalgebraPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The condition that 1/4T and 2/7T must both be whole numbers forces T to be a multiple of lcm(4,7)=28, so the candidate totals are just T=28,56,84,… — a tiny list. Tool #6 (Guess and Check) on those candidates is the fastest path: for each candidate compute Alexa + Brittany + Chelsea, subtract from T, and keep the one whose leftover X obeys 0 ≤ X ≤ 14. Tool #3 (Eliminate) supports it by ruling out totals where X would be negative or above 14. Tool #7 (Identify Subproblems) sets up the bookkeeping: "score budget", "three named players", "seven others".
Write the score budget
Split the score into three named players plus the seven others; the four pieces must sum to T.
Taking a fraction of a whole quantity (1/4 of T, 2/7 of T) is the Grade 5 "fraction of a whole" word-problem move.
5.NF.B.6Identify SubproblemsMake the shares whole numbers
Whole-number scores force T and T to be integers, so T must be divisible by 4 and 7 — a multiple of 28.
Finding the least common multiple of 4 and 7 to combine the divisibility requirements is Grade 6 LCM reasoning.
The team's total number of points T can only be a whole-number total if it is a multiple of 28.
▸ Why?
Alexa's quarter and Brittany's two-sevenths both come out as whole scores, so T has to be a multiple of 4 and a multiple of 7 at the same time; the first total that is both is 28, and every total after it is another whole 28 further on.
▸ Why?
T is a multiple of 4: Alexa's share is one quarter of T and a score must be a whole number a, so T is just four of those equal quarters stacked back together, T = 4 × a — four equal groups of a, which is what being a multiple of 4 means.
▸ Why?
T is a multiple of 7: Brittany's two-sevenths of T is a whole score, so the points must fall into seven equal whole shares s, and T is seven of those shares, T = 7 × s — seven equal groups, which is what being a multiple of 7 means.
▸ Why?
A total that is a multiple of 4 and a multiple of 7 at once is a common multiple of 4 and 7, and the common multiples of two numbers are exactly the multiples of their least common multiple; since 4 and 7 share no smaller common factor that LCM is 4 × 7 = 28, so T first works at 28 and then steps by 28.
Bound the total score
The seven others get T − 15 points and can total at most 14, so T − 15 ≤ 14 gives T ≤ 62.
Translating "each of 7 scores at most 2" into an inequality on T is exactly the Grade 6 inequality-from-a-real-world-constraint standard.
6.EE.B.8Eliminate PossibilitiesCombine the two filters
Intersecting multiple-of-28 with T ≤ 62 leaves only T = 28 or T = 56 to test by hand.
Intersecting "multiple of 28" with the upper bound leaves a tiny candidate set — a Grade 6 multiples-and-bounds check.
6.NS.B.4Guess And CheckTest a total of 28
Try T = 28: Alexa 7 + Brittany 8 + Chelsea 15 = 30 already tops 28, forcing X = -2 — impossible, so reject.
Computing 1/4 and 2/7 of a whole number is a Grade 5 fraction-of-a-whole calculation.
5.NF.B.6Guess And CheckTest a total of 56
Try T = 56: Alexa 14 + Brittany 16 + Chelsea 15 = 45, so the seven others scored X = 11, and 0 ≤ 11 ≤ 14 checks out.
Adding and subtracting whole numbers to compute the leftover score is a Grade 4 multi-step word-problem skill.
4.OA.A.3Guess And CheckThis AMC 8 problem only needs Grade 6 least-common-multiple reasoning you already know — T has to be a multiple of both 4 and 7, so it jumps in steps of 28!
- Write the score budget
- Make the shares whole numbers
- Bound the total score
- Combine the two filters
- Test a total of 28
- Test a total of 56
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