AMC 8 · 2019 · #23

Grade 6 number-theoryalgebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen

lcmmultiplesfraction-arithmeticlinear-equations-one-var convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: lcmfraction-arithmetic

📏 Long solution 💡 3 insights

Problem

Euclid High School's team has a total score T. Alexa scored

\frac{1}{4}

T, Brittany scored

\frac{2}{7}

T, and Chelsea scored 15. The remaining 7 teammates each scored at most 2 points. Find the total X scored by those 7 teammates.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

The condition that $\frac{1}{4}$ T and $\frac{2}{7}$ T 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".

1STEP 1

Split the score into three named players plus the seven others; the four pieces must sum to T.

\frac{1}{4}

T_Alexa +

\frac{2}{7}

T_Brittany + 15_Chelsea + X₇ others = T

2STEP 2

Whole-number scores force $\frac{1}{4}$ T and $\frac{2}{7}$ T to be integers, so T must be divisible by 4 and 7 — a multiple of 28.

T ∈ {28, 56, 84, 112,…}

3STEP 3

The seven others get $\frac{13}{28}$ T − 15 points and can total at most 14, so $\frac{13}{28}$ T − 15 ≤ 14 gives T ≤ 62.

\frac{13}{28}

T - 15 ≤ 14 ⟹ T ≤ 62.46…

4STEP 4

Intersecting multiple-of-28 with T ≤ 62 leaves only T = 28 or T = 56 to test by hand.

Candidates for T: {28, 56}

5STEP 5

Try T = 28: Alexa 7 + Brittany 8 + Chelsea 15 = 30 already tops 28, forcing X = -2 — impossible, so reject.

T=28 → X = 28 - (7+8+15) = -2 ＜ 0 — reject.

6STEP 6

Try T = 56: Alexa 14 + Brittany 16 + Chelsea 15 = 45, so the seven others scored X = 11, and 0 ≤ 11 ≤ 14 checks out.

T=56 → X = 56 - (14+16+15) = 11 → (B)

Answer

Cross-check that X=11 can actually happen: 7 players each scoring at most 2 can hit 11 total, e.g. four players score 2 and three players score 1 (4 × 2 + 3 × 1 = 11). The final tallies 14+16+15+11=56=T match perfectly. The next candidate, T=84, would force the 7 others to score

\frac{13}{28}

(84)-15 = 39-15 = 24 ＞ 14, which is impossible — confirming T=56 is unique.

💡Key takeaway

This 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!

More like this