Difference between revisions of "2017 AMC 10A Problems/Problem 16"
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<math>\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6</math> | <math>\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6</math> | ||
| − | ==Solution== | + | ==Solution 1== |
If we have horses, <math>a_1, a_2, \ldots, a_n</math>, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that <math>\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12</math>. Finally, <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>. | If we have horses, <math>a_1, a_2, \ldots, a_n</math>, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that <math>\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12</math>. Finally, <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We are trying to find the smallest number that has <math>5</math> one-digit divisors. We quickly consider <math>12</math> since it is the smallest number that has at least <math>5</math> divisors. Since <math>12</math> has <math>5</math> single-digit divisors, namely <math>1</math>,<math>2</math>,<math>3</math>,<math>4</math>,<math>6</math>, our answer is <math>1+2 = \boxed{\textbf{(B)}\ 3}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 06:52, 9 February 2017
Contents
Problem
There are 
horses, named Horse 
, Horse 
, . . . , Horse . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse 
runs one lap in exactly minutes. At time 
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 
, in minutes, at which all horses will again simultaneously be at the starting point is 
. Let 
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 
Solution 1
If we have horses, 
, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that 
. Finally, 
.
Solution 2
We are trying to find the smallest number that has 
one-digit divisors. We quickly consider 
since it is the smallest number that has at least divisors. Since has single-digit divisors, namely ,,
,
,
, our answer is
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
