AMC 10 · 2017 · #16
Easy mode Grade 4There are 10 horses on a circular track, named Horse 1 through Horse 10. Horse k takes exactly k minutes to run one lap, so Horse 1 takes 1 minute, Horse 2 takes 2 minutes, and so on. At time 0 all the horses start together at the starting point and keep running at their own steady speeds.
The first time all 10 horses are back at the starting point together is at S=2520 minutes. Now let T be the earliest time, in minutes, when at least 5 of the horses are back at the starting point at the same moment. What is the sum of the digits of T?
Pick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $T$.
Givens: There are $10$ horses; Horse $k$ takes exactly $k$ minutes per lap, for $k=1,2,\dots,10$; All horses leave the starting point together at time $0$ and keep constant speeds; We are told the time when all $10$ meet again is $S=2520$ minutes; $T$ is the least time when at least $5$ horses are at the start at once; Answer choices are the digit sum of $T$: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The least time $T>0$ when $5$ or more horses are all at the start; The sum of the digits of that $T$
Understand
Restated: 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 $T$.
Givens: There are $10$ horses; Horse $k$ takes exactly $k$ minutes per lap, for $k=1,2,\dots,10$; All horses leave the starting point together at time $0$ and keep constant speeds; We are told the time when all $10$ meet again is $S=2520$ minutes; $T$ is the least time when at least $5$ horses are at the start at once; Answer choices are the digit sum of $T$: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #6 Guess and Check
Secondary: #4 Introduce a Variable, #2 Make a Systematic List
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 $T=1,2,3,\dots$ 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.
Execute — Answer: B
4.OA.B.4 Step 1 Turn position into divisibility
- Horse $k$ runs one lap in $k$ minutes, so it is back at the starting point at minutes $k, 2k, 3k, \dots$ — every multiple of $k$.
- Therefore at time $T$, Horse $k$ stands at the start exactly when $T$ is a multiple of $k$, that is, when $k$ divides $T$.
- Counting how many horses are at the start at time $T$ is the same as counting how many of the numbers $1,2,\dots,10$ divide $T$.
💡 A horse only meets the start line at whole numbers of its own laps, so its return times are exactly the multiples of $k$.
4.OA.A.3 Step 2 Restate the goal
- "At least $5$ horses at the start at time $T$" now means $T$ is divisible by at least $5$ of the numbers $1$ through $10$ — in other words, $T$ has at least $5$ divisors that lie in the range $1$ to $10$.
- We want the smallest positive $T$ with this property.
- Note we need only $5$ horses, not all $10$, so $T$ should be far smaller than the all-meet time $S=2520$; that means we can safely search upward from small numbers.
💡 Asking for $5$ horses is asking $T$ to carry $5$ small divisors at once — a question purely about factors.
4.OA.C.5 Step 3 Search upward and count divisors
- Test $T=1,2,3,\dots$ and for each count its divisors from $1$ to $10$.
- Small numbers stall at four: $T=6$ has $\{1,2,3,6\}$, $T=8$ has $\{1,2,4,8\}$, and $T=10$ has $\{1,2,5,10\}$ — each only four.
- The first jump to five comes at $T=12$, whose divisors in range are $\{1,2,3,4,6\}$ (the number $12$ itself is past Horse $10$, so it does not count).
- Since every $T$ below $12$ tops out at four, $T=12$ is the least time with at least $5$ horses at the start.
💡 Climb the numbers one at a time; the first one rich enough in small factors to feed five horses is the winner.
4.OA.A.3 Step 4 Add the digits of $T$
- The least meeting time for $5$ horses is $T=12$ minutes.
- Its digits are $1$ and $2$, and their sum is $1+2=3$.
- So the sum of the digits of $T$ is $3$, which is choice (B).
💡 Once you have the time itself, the digit sum is just a quick final addition.
4.OA.B.4 Horse $k$ runs one lap in $k$ minutes, so it is back at the starting point at mi 4.OA.A.3 "At least $5$ horses at the start at time $T$" now means $T$ is divisible by at 4.OA.C.5 Test $T=1,2,3,\dots$ and for each count its divisors from $1$ to $10$. Small num 4.OA.A.3 The least meeting time for $5$ horses is $T=12$ minutes. Its digits are $1$ and Review
Reasonableness: 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).
Alternative: Instead of scanning every number, look for the cheapest number that packs in five small divisors — a highly composite number. The all-divisor counts grow as $1,2,4,6,12,\dots$: the number $12$ is the first whole number with six total divisors ($1,2,3,4,6,12$), five of which fall in the range $1$ to $10$. No smaller number reaches five single-digit-range divisors, so $T=12$ and the digit sum is $1+2=3$, again (B).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Translating "Horse $k$ returns every $k$ minutes" into the divisibility statement $k \mid T$, and counting which of $1$–$10$ divide a given $T$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Restating the "at least $5$ horses" condition as a divisor count and then adding the digits of $T$ to finish.)4.OA.C.5Generate a number or shape pattern following a given rule (Stepping through $T=1,2,3,\dots$ and tracking how many qualifying divisors each value has until five first appears.)
⭐ 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 $3$.
⭐ 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 $3$.
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