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Difference between revisions of "2013 AIME II Problems/Problem 7"

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There are <math>x</math> clerks at the beginning, and <math>t</math> clerks are reassigned to another task at the end of each hour. So, <math>30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775</math>, and simplify that we get <math>19x-21t=355</math>.
 
There are <math>x</math> clerks at the beginning, and <math>t</math> clerks are reassigned to another task at the end of each hour. So, <math>30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775</math>, and simplify that we get <math>19x-21t=355</math>.
 
Now the problem is to find a reasonable integer solution. Now we know <math>x= \frac{355+21t}{19}</math>, so <math>19</math> divides <math>355+21t</math>, AND as long as <math>t</math> is a integer, <math>19</math> must divide <math>2t+355</math>. Now, we suppose that <math>19m=2t+355</math>, similarly we get <math>t=\frac{19m-355}{2}</math>, and so in order to get a minimum integer solution for <math>t</math>, it is obvious that <math>m=19</math> works. So we get <math>t=3</math> and <math>x=22</math>.  One and a half hour's work should be <math>30x+15(x-t)</math>, so the answer is <math>\boxed{945}</math>.
 
Now the problem is to find a reasonable integer solution. Now we know <math>x= \frac{355+21t}{19}</math>, so <math>19</math> divides <math>355+21t</math>, AND as long as <math>t</math> is a integer, <math>19</math> must divide <math>2t+355</math>. Now, we suppose that <math>19m=2t+355</math>, similarly we get <math>t=\frac{19m-355}{2}</math>, and so in order to get a minimum integer solution for <math>t</math>, it is obvious that <math>m=19</math> works. So we get <math>t=3</math> and <math>x=22</math>.  One and a half hour's work should be <math>30x+15(x-t)</math>, so the answer is <math>\boxed{945}</math>.
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==Solution 2==
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We start with the same approach as solution 1 to get <math>19x-21t=355</math>. Then notice that <math>21t + 355 \equiv 0 \pmod{19}</math>, or <math>2t-6 \equiv 0 \pmod{19}</math>, giving the smallest solution at <math>t=3</math>. We find that <math>x=22</math>. Then the number of files they sorted will be <math>30x+15(x-t)=660+285=\boxed{945}.</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2013|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:52, 7 March 2022

Problem 7

A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

Solution

There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775$, and simplify that we get $19x-21t=355$. Now the problem is to find a reasonable integer solution. Now we know $x= \frac{355+21t}{19}$, so $19$ divides $355+21t$, AND as long as $t$ is a integer, $19$ must divide $2t+355$. Now, we suppose that $19m=2t+355$, similarly we get $t=\frac{19m-355}{2}$, and so in order to get a minimum integer solution for $t$, it is obvious that $m=19$ works. So we get $t=3$ and $x=22$. One and a half hour's work should be $30x+15(x-t)$, so the answer is $\boxed{945}$.

Solution 2

We start with the same approach as solution 1 to get $19x-21t=355$. Then notice that $21t + 355 \equiv 0 \pmod{19}$, or $2t-6 \equiv 0 \pmod{19}$, giving the smallest solution at $t=3$. We find that $x=22$. Then the number of files they sorted will be $30x+15(x-t)=660+285=\boxed{945}.$

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions

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