2017 AMC 12A Problem 12

Question

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\gt0,

in minutes, at which all

10

horses will again simultaneously be at the starting point is

S=2520.

Let

T\gt0

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?

2

3

4

5

6

Answer: B

Concepts:least common multiple divisibility

Difficulty rating: 1630

Small hint:

Horse

k

is back at the start at time

t

exactly when

k

divides

t

Big hint:

Find the smallest

t

having at least

5

divisors among

1

through

10

Solution:

Horse

k

is at the starting point at time

t

precisely when

t

is a multiple of

k.

So we want the smallest

t

with at least

5

divisors among

1,2,\ldots,10.

The positive integers below

12

have at most

4

divisors, while

12

is divisible by

1,2,3,4,

and

6.

Thus

T=12,

and the sum of its digits is

1+2=3.

Thus, the correct answer is B.

2017 AMC 12A Problem 12

Answer: B

Difficulty rating: 1630

Small hint:

Big hint:

Solution:

Problem 12 in other years