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>, 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>, 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2013|n=II|num-b=6|num-a=8}} |
Revision as of 16:37, 6 April 2013
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
, 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
.
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 |