Difference between revisions of "1952 AHSME Problems/Problem 38"
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== Solution == | == Solution == | ||
| − | Let us denote <math>8m</math> and <math>8n</math> to be our bases. Without loss of generality, <math>m | + | Let us denote <math>8m</math> and <math>8n</math> to be our bases. Without loss of generality, <math>m \le n</math> |
Thus, | Thus, | ||
Revision as of 00:18, 22 December 2015
Problem
The area of a trapezoidal field is 
square yards. Its altitude is 
yards. Find the two bases, if the number of yards in each base is an integer divisible by 
. The number of solutions to this problem is:
Solution
Let us denote 
and 
to be our bases. Without loss of generality, 
Thus,
![\[50 * \frac{8m + 8n}{2} = 1400\]](http://latex.artofproblemsolving.com/1/4/7/1471e500538e266a6ef73ee9fe4ce4a05521aff5.png)
![\[4m + 4n = 28\]](http://latex.artofproblemsolving.com/c/f/2/cf290b7e13ff6e12abac868b2b6afd21d6ceff45.png)
![\[m + n = 7\]](http://latex.artofproblemsolving.com/4/f/0/4f0c72d03b86a57185c4c9991cd1230fe52eee4f.png)
Since 
& 
are integers, we see that the only solutions to this equation are 
, 
, and 
.
Therefore, the answer is 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 37 |
Followed by Problem 39 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
