Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #23
Grade 4 rate-rationumber-theoryPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Write the daily distance formula
In 60 minutes at m minutes per mile, Linda covers 60/m miles — so m must be a factor of 60.
Dividing minutes by minutes-per-mile leaves miles — a Grade 4 distance/time word-problem move.
4.MD.A.2Analyze The UnitsList the factors of 60
List every factor of 60 in order — these are the only legal values for the four per-mile times.
Writing out every factor pair of 60 is Grade 4 factor-finding, and a systematic list guarantees we miss nothing.
4.OA.B.4Make A Systematic ListCheck the candidate paces
Test m₁ = 1, 2, 3, … in order; the paces jump by 5, and only m₁ = 5 lands all four on factors of 60.
Building a sequence with a fixed +5 rule and testing whether each term lands on a factor of 60 is exactly Grade 4 "generate a number pattern following a rule."
The only first-day per-mile time that keeps all four days at a whole number of miles is 5 minutes, which forces the four daily paces to be 5, 10, 15, and 20 minutes per mile.
▸ Why?
Each day's mileage is the fixed 60 minutes of walking divided by that day's per-mile pace, so a day lands on whole miles only when its pace divides 60 with no minutes left over.
▸ Why?
Walking for one hour is the same 60 minutes every day, so each day starts from the identical fixed 60 minutes to hand out mile by mile.
▸ Why?
Sharing 60 minutes into equal per-mile pieces is the undo of building 60 back up from equal pieces, so it comes out whole only when the pace fits into 60 a whole number of times.
▸ Why?
Because the pace climbs by a fixed 5 minutes each day, the four paces are four divisors of 60 spaced 5 apart, and sliding that window of four up the divisor list 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 leaves 5, 10, 15, 20 as the one run that stays entirely on the list — so the first pace has to be 5.
▸ Why?
5, 10, 15, and 20 each divide 60 evenly, since 60 is twelve 5s, six 10s, four 15s, and three 20s, so all four of those days give whole miles.
Convert each pace to distance
With m₁ = 5 the paces 5, 10, 15, 20 min/mile give daily distances 12, 6, 4, 3 miles.
Dividing 60 by single- and two-digit factors uses Grade 3 fluent multiplication/division facts within 100.
3.OA.C.7Analyze The UnitsAdd the four distances
Add the four daily distances: 12 + 6 + 4 + 3 = 25 miles → (C).
Adding four small whole numbers fluently is the Grade 4 multi-digit addition standard.
4.NBT.B.4Analyze The UnitsThis AMC 8 problem only needs Grade 4 factor-finding and pattern-making you already know!
- Write the daily distance formula
- List the factors of 60
- Check the candidate paces
- Convert each pace to distance
- Add the four distances
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