AMC 10 · 2017 · #16

Grade 4 rate-ratio
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
lcmdivisor-countmultiples systematic-enumeration ↑ Prerequisites: lcm
📏 Medium solution 💡 3 insights
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Problem
Ten horses run laps; Horse k finishes one lap every k minutes, so all start together at time 0. At time T a horse is back at the starting point only when T is a whole number of its laps. Find the least time T ＞ 0 at which at least 5 of the horses are simultaneously at the start, then add up the digits of T.

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
5
(E)
6

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

The hidden engine of this problem is one translation: Horse k returns to the start every k minutes, so at time T it stands at the start exactly when k divides T. Tool #4 (Introduce a Variable) lets us name the meeting time T and restate the whole race as a divisibility condition. Once that is clear, "at least 5 horses at the start" becomes "T has at least 5 divisors among 1 through 10," and the smallest such T is easy to hunt for by Tool #6 (Guess and Check): test T=1,2,3,… in order and count qualifying divisors, using Tool #2 (Make a Systematic List) to keep the tally tidy. Because we only need 5 of the 10 horses, T will be far below the all-meet time S=2520, so checking small numbers by hand is quick.

1STEP 1

Turn position into divisibility

Horse k returns to the start at every multiple of k, so at time T it is there exactly when k divides T.

Horse k at start at time T ⇔ k ∣ T
2STEP 2

Restate the goal

'At least 5 horses at time T' means T has at least 5 divisors in 1–10; find the smallest such T, far below the all-meet time 2520.

find least T ＞ 0 with #{k ∈ {1,…,10} : k ∣ T} ≥ 5
3STEP 3

Search upward and count divisors

Climbing upward, counts stall at four (6,8,10) until T=12, whose in-range divisors {1,2,3,4,6} are five — the first to reach five.

12: {1,2,3,4,6} (5 divisors) vs. 6,8,10: 4 each
4STEP 4

Add the digits of T

T=12 has digits 1 and 2, so the digit sum is 1+2=3 — choice (B).

T=12 → 1+2=3 → (B)
Answer
3
The answer T=12 is much smaller than the all-meet time S=2520, exactly as expected: gathering only 5 horses is far easier than gathering all 10, so the time should be far shorter. A direct check confirms 12 works — Horses 1,2,3,4,6 are all at the start at minute 12 — and a sweep of every number below 12 shows none reaches five qualifying divisors (the best are 6, 8, and 10 with four each). The digits 1 and 2 sum to 3, matching choice (B).
💡Key takeaway

A horse is back at the start whenever the clock hits a multiple of its number, so finding when 5 horses meet is just finding the smallest time with 5 small divisors — that is 12, and its digits add to 3.

  • Turn position into divisibility
  • Restate the goal
  • Search upward and count divisors
  • Add the digits of T

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