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2017 AMC 10A Problem 16

2017 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 3 min · position P16-20

There are 10 horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S>0$ , in minutes, at which all 10 horses will again simultaneously be at the starting point is $S=2520$ . Let $T>0$ be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of $T$ ?

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

Answer

Key insight

Horse k is at the start at time T exactly when k divides T, so T is the least number with five divisors among 1 to 10: T = 12.

Solution

Horse $k$ is back at the starting point at time $t$ exactly when $t$ is a multiple of $k$ . So at time $T$ the horses at the start are the horses whose numbers divide $T$ , and we want the smallest $T$ with at least five divisors in $\{1, 2, \ldots, 10\}$ .

Any $T \le 11$ has at most four divisors in total ( $6$ , $8$ and $10$ each have exactly four; the primes have two), so none works.

$T = 12$ has divisors $1, 2, 3, 4, 6, 12$ , and five of them are at most $10$ : Horses $1, 2, 3, 4, 6$ are all at the start after $12$ minutes.

So $T = 12$ and the sum of its digits is $1 + 2 = 3$ .

The answer is $\boxed{\textbf{(B)}\ 3}$ .

Why this works

The story about laps is a disguise for divisibility: "Horse $k$ is at the start" means " $k \mid T$ ." The given value $S = 2520 = \text{lcm}(1, \ldots, 10)$ is the "all horses" version; the "at least five" version asks for the smallest highly divisible number instead, and $12$ is the smallest positive integer with more than four divisors.

The trap

Reading 'at least 5 horses' as 'Horses 1 through 5' and computing lcm(1,...,5) = 60, whose digit sum 6 is choice (E).

Common mistakes

Reading 'at least 5 horses' as 'Horses 1 through 5' and computing lcm(1,...,5) = 60, whose digit sum 6 is choice (E).
Answering $T = 12$ itself or forgetting the final step of summing the digits.

Techniques

Bound the quantity above/below or estimate to pin it down · Organized listing / direct enumeration