Difference between revisions of "2013 AIME II Problems/Problem 7"
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==Solution== | ==Solution== | ||
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>, 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
Contents
Problem 7
A group of clerks is assigned the task of sorting 
files. Each clerk sorts at a constant rate of 
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 
hours and 
minutes. Find the number of files sorted during the first one and a half hours of sorting.
Solution
There are 
clerks at the beginning, and 
clerks are reassigned to another task at the end of each hour. So, 
, and simplify that we get 
.
Now the problem is to find a reasonable integer solution. Now we know 
, so 
divides 
, AND as long as is a integer, must divide 
. Now, we suppose that 
, similarly we get 
, and so in order to get a minimum integer solution for , it is obvious that 
works. So we get 
and 
. One and a half hour's work should be 
, so the answer is 
.
Solution 2
We start with the same approach as solution 1 to get . Then notice that 
, or 
, giving the smallest solution at . We find that . Then the number of files they sorted will be 
See Also
| 2013 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
