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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.
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Convert minutes to hours by again, dividing by <math>60</math>: <math>9\cdot 8\cdot 7\cdot 2</math> hours.
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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.
  
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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 10}}</math>
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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 and Max0815
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~nosysnow ~Max0815
  
 
==Video Solutions==
 
==Video Solutions==

Revision as of 03:38, 5 January 2022

Problem

A unit of blood expires after $10!=10\cdot 9 \cdot 8 \cdots 1$ seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?

$\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}$

Solution

The problem says there are $10! = 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ seconds. Convert $10!$ seconds to minutes by dividing by $60$: $9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2$ minutes. Convert minutes to hours by dividing by $60$ again: $9\cdot 8\cdot 7\cdot 2$ hours. Convert hours to days by dividing by $24$: $3\cdot 7\cdot 2 = 42$ days.

Now we need to count $42$ 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. $42 - 30 = 12$. Count $12$ more days, resulting in $\fbox{\textbf{(E) }\text{February 12}}$.

~nosysnow ~Max0815

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/FbSYTL8tPwo

~savannahsolver

https://youtu.be/bPfLeXu9kx0

Education, the Study of Everything

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. AMC logo.png

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