AMC 8 · 2018 · #13

Grade 6 arithmeticalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

mean-median-mode-rangelinear-equations-one-varbound-inequality-then-enumeratesystematic-enumeration bound-inequality-then-enumeratesystematic-enumerationconvert-to-algebra ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

Laila took 5 tests, each scored on integers from 0 to 100. She got the SAME score on tests 1 through 4 (call it x) and a STRICTLY HIGHER score on test 5 (call it y, with y ＞ x). Her average across the 5 tests was 82. How many different integer values are possible for y?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

Once we turn "average = 82" into the total 4x + y = 410, the unknown y is squeezed into a small window (just above 82, at most 100). That is a tiny, bounded set — perfect for Tool #6 (Guess and Check): try each candidate y and see whether the forced x = (410 - y)/4 is a whole number that is less than y. Tool #2 (Systematic List) keeps the candidate scan in strict order (smallest y first) so nothing is missed or repeated, and Tool #3 (Eliminate Possibilities) is how we drop any y whose forced x is not an integer or violates y ＞ x.

1STEP 1

Turn the average into a total: 5 × 82 = 410, and since four scores are x and one is y, that means 4x + y = 410.

\frac{x + x + x + x + y}{5}

= 82 ⟹ 4x + y = 410

2STEP 2

Since y ＞ x, y must beat the average, so y ＞ 82; with the cap y ≤ 100 that traps 83 ≤ y ≤ 100.

y ＞ 82 and y ≤ 100 ⟹ 83 ≤ y ≤ 100

3STEP 3

Solve 4x = 410 − y: x is whole only when 410 − y is a multiple of 4, and since 410 = 4 · 102 + 2, y must leave remainder 2 mod 4.

4x = 410 - y ⟹ y leaves remainder 2 when divided by 4

4STEP 4

List y in 83–100 leaving remainder 2: starting at 86 and adding 4 gives 86, 90, 94, 98 — the next, 102, exceeds 100.

y ∈ {86, 90, 94, 98}

5STEP 5

Check each: x = (410 − y)/4 gives 81, 80, 79, 78, all with y ＞ x, so all four survive — 4 values.

y = 86 → x = 81, y = 90 → x = 80, y = 94 → x = 79, y = 98 → x = 78 — all integers with y ＞ x. Count = 4 → (A)

Answer

Sanity check the extremes. The smallest legal y is 86 with x = 81: average =

\frac{4 · 81 + 86}{5}

\frac{410}{5}

= 82 and 86 ＞ 81. The largest legal y is 98 with x = 78: average =

\frac{4 · 78 + 98}{5}

\frac{410}{5}

= 82 and 98 ＞ 78. Both ends work, and the four candidates step by 4 (since each unit increase in y must be matched by

\frac{1}{4}

unit drop in x, which requires y to change by multiples of 4 to keep x integer). Four values matches choice (A).

💡Key takeaway

This AMC 8 problem only needs Grade 6 mean-as-a-total reasoning plus a quick inequality you already know!

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