Difference between revisions of "2018 AMC 10A Problems/Problem 3"
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<math>\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}</math> | <math>\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}</math> | ||
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== Solution == | == Solution == | ||
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The problem says there are <math>10! = 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1</math> seconds. | The problem says there are <math>10! = 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1</math> seconds. | ||
Convert <math>10!</math> seconds to minutes by dividing by <math>60</math>: <math>9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2</math> minutes. | Convert <math>10!</math> seconds to minutes by dividing by <math>60</math>: <math>9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2</math> minutes. | ||
| − | Convert minutes to hours by | + | Convert minutes to hours by dividing by <math>60</math> again: <math>9\cdot 8\cdot 7\cdot 2</math> hours. |
Convert hours to days by dividing by <math>24</math>: <math>3\cdot 7\cdot 2 = 42</math> days. | Convert hours to days by dividing by <math>24</math>: <math>3\cdot 7\cdot 2 = 42</math> days. | ||
| − | Now we need to count <math>42</math> days after January 1. Since we start on Jan. 1, then we can't count that as a day itself. When we reach Jan. 31(The end of the month), we only have counted 30 days. <math>42 - 30 = 12</math>. Count <math>12</math> more days, resulting <math>\fbox{\textbf{(E) }\text{February 12}}</math> | + | Now we need to count <math>42</math> days after January 1. Since we start on Jan. 1, then we can't count that as a day itself. When we reach Jan. 31(The end of the month), we only have counted 30 days. <math>42 - 30 = 12</math>. Count <math>12</math> more days, resulting in <math>\fbox{\textbf{(E) }\text{February 12}}</math>. |
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| + | ~nosysnow ~Max0815 | ||
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| + | ==Solution 2 (Concise)== | ||
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| + | There are <math>60 \cdot 60 \cdot 24 = 86400</math> seconds in a day, which means that Yasin's blood expires in <math>10! \div 86400 = 42</math> days. Since there are <math>31</math> days in January (consult a calendar), then <math>42-31+1</math> (Jan 1 doesn't count) is <math>12</math> days into February, so <math>\boxed{\textbf{(E) }\text{February 12}}</math>. | ||
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| + | ~MrThinker | ||
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| + | ==Video Solution (HOW TO THINK CREATIVELY)== | ||
| + | https://youtu.be/bPfLeXu9kx0 | ||
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| + | Education, the Study of Everything | ||
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==Video Solutions== | ==Video Solutions== | ||
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~savannahsolver | ~savannahsolver | ||
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== See Also == | == See Also == | ||
Latest revision as of 14:37, 3 July 2023
Contents
Problem
A unit of blood expires after 
seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?
Solution
The problem says there are 
seconds.
Convert 
seconds to minutes by dividing by 
: 
minutes.
Convert minutes to hours by dividing by again: 
hours.
Convert hours to days by dividing by 
: 
days.
Now we need to count 
days after January 1. Since we start on Jan. 1, then we can't count that as a day itself. When we reach Jan. 31(The end of the month), we only have counted 30 days. 
. Count 
more days, resulting in 
.
~nosysnow ~Max0815
Solution 2 (Concise)
There are 
seconds in a day, which means that Yasin's blood expires in 
days. Since there are 
days in January (consult a calendar), then 
(Jan 1 doesn't count) is days into February, so 
.
~MrThinker
Video Solution (HOW TO THINK CREATIVELY)
Education, the Study of Everything
Video Solutions
~savannahsolver
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
