Difference between revisions of "2017 AMC 10A Problems/Problem 16"
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==Problem== | ==Problem== | ||
− | There are <math>10</math> horses, named Horse <math>1</math>, Horse <math>2</math>, . . . , Horse <math>10</math>. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse <math>k</math> runs one lap in exactly <math>k</math> minutes. At time <math>0</math> 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 <math>S > 0</math>, in minutes, at which all <math>10</math> horses will again simultaneously be at the starting point is <math>S=2520</math>. Let <math>T > 0</math> 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 <math>T?</math> | + | There are <math>10</math> horses, named Horse <math>1</math>, Horse <math>2</math>, . . . , Horse <math>10</math>. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse <math>k</math> runs one lap in exactly <math>k</math> minutes. At time <math>0</math> 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 <math>S > 0</math>, in minutes, at which all <math>10</math> horses will again simultaneously be at the starting point is <math>S=2520</math>. Let <math>T > 0</math> be the least time, in minutes, such that at least <math>5</math> of the horses are again at the starting point. What is the sum of the digits of <math>T?</math> |
<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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We are trying to find the smallest number that has <math>5</math> one-digit divisors. Therefore we try to find the LCM for smaller digits, such as <math>1</math>,<math> 2</math>, <math>3</math>, or <math>4</math>. We quickly consider <math>12</math> since it is the smallest number that is the LCM of <math>1</math>, <math>2</math>, <math>3</math> and <math>4</math>. Since <math>12</math> has <math>5</math> single-digit divisors, namely <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>6</math>, our answer is <math>1+2 = \boxed{\textbf{(B)}\ 3}</math> | ||
+ | |||
+ | ==Solution 3 (Speedy Guess and Check)== | ||
+ | First, for 5 horses to simultaneously pass the starting line after <math>T</math> seconds, <math>T</math> must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of <math>T</math>, and therefore meaning <math>T</math> must have at least <math>5</math> <math>1</math>-digit divisors. Since we want to minimize <math>T</math>, we will start by guessing the lowest natural number, <math>1</math>. <math>1</math> has only <math>1</math> factor, so it does not work, we now repeat the process for the numbers between <math>2</math> and <math>12</math> (This should not take more than a minute) to get that <math>12</math> is the first number to have <math>5</math> or more single-digit divisors (<math>1, 2, 3, 4, 6</math>). The sum of the digits of <math>12</math> is <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>, which is our answer. | ||
+ | |||
+ | Note that this solution is rather fast with this problem because the numbers given in the question are low, this may not always be the case, however, when given higher numbers. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/umr2Aj9ViOA | ||
==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}} | ||
+ | {{AMC12 box|year=2017|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 23:20, 20 January 2020
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 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. Therefore we try to find the LCM for smaller digits, such as ,, , or . We quickly consider since it is the smallest number that is the LCM of , , and . Since has single-digit divisors, namely , , , , and , our answer is
Solution 3 (Speedy Guess and Check)
First, for 5 horses to simultaneously pass the starting line after seconds, must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of , and therefore meaning must have at least -digit divisors. Since we want to minimize , we will start by guessing the lowest natural number, . has only factor, so it does not work, we now repeat the process for the numbers between and (This should not take more than a minute) to get that is the first number to have or more single-digit divisors (). The sum of the digits of is , which is our answer.
Note that this solution is rather fast with this problem because the numbers given in the question are low, this may not always be the case, however, when given higher numbers.
Video Solution
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
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