AMC 8 · 2017 · #23

Grade 4 rate-rationumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

ratefactorssequences-arithmeticdivisibility-rules systematic-enumerationguess-and-check ↑ Prerequisites: ratefactorssequences-arithmetic

📏 Medium solution 💡 3 insights

Problem

For four straight days Linda walks for exactly 1 hour (= 60 minutes) each day. On Day 1 it takes her a whole number of minutes, m₁, to cover one mile. Every following day that per-mile time grows by 5 minutes (so Day 2 takes m₁+5 min/mile, Day 3 takes m₁+10, Day 4 takes m₁+15). On every day the total miles she covers is also a whole number. Find the total miles she walks over the four days.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

The integer-distance condition forces every per-mile time to be a factor of 60. There are only 12 such factors, so we use Tool #2 (Systematic List) to write them all out, then Tool #6 (Guess and Check) to test each candidate value of m₁ in order: does m₁, m₁+5, m₁+10, m₁+15 keep landing on factors of 60? Tool #8 (Analyze the Units) underpins the setup — the relationship distance = (60 min)/(m min/mile) has units of miles, which is exactly what the problem asks for.

1STEP 1

In 60 minutes at m minutes per mile, Linda covers 60/m miles — so m must be a factor of 60.

d_k = (60 min)/(m_k min/mile) = 60/m_k miles

2STEP 2

List every factor of 60 in order — these are the only legal values for the four per-mile times.

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

3STEP 3

Test m₁ = 1, 2, 3, … in order; the paces jump by 5, and only m₁ = 5 lands all four on factors of 60.

m₁=1 : (1,6,11,16) ✗ m₁=2 : (2,7,12,17) ✗ m₁=3 : (3,8,13,18) ✗ m₁=4 : (4,9,14,19) ✗ m₁=5 : (5,10,15,20) ✓

4STEP 4

With m₁ = 5 the paces 5, 10, 15, 20 min/mile give daily distances 12, 6, 4, 3 miles.

d₁=

\frac{60}{5}

=12, d₂=

\frac{60}{10}

=6, d₃=

\frac{60}{15}

=4, d₄=

\frac{60}{20}

5STEP 5

Add the four daily distances: 12 + 6 + 4 + 3 = 25 miles → (C).

12 + 6 + 4 + 3 = 25 miles → (C)

Answer

Sanity-check the four daily distances. At 5 min/mile Linda is jogging quickly (12 mph) for 12 miles; at 20 min/mile she is walking (3 mph) for 3 miles. The distances 12, 6, 4, 3 drop off as the pace slows, which matches the story. Their sum, 25, is choice (C), and the only other near-by choice (50) would require doubling everything — but m₁ = 10 gives the sequence (10, 15, 20, 25), and 25 is not a factor of 60, so no second solution exists. The answer is uniquely (C).

💡Key takeaway

This AMC 8 problem only needs Grade 4 factor-finding and pattern-making you already know!

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