Difference between revisions of "1952 AHSME Problems/Problem 28"
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Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7. | Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7. | ||
Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31. | Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31. | ||
| − | Thus, | + | Thus, the answer is <math>\boxed{\bf{(C)}}</math>. |
== See also == | == See also == | ||
Revision as of 01:10, 13 November 2023
Problem
In the table shown, the formula relating x and y is:
![\[\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\ \hline y & 3 & 7 & 13 & 21 & 31\\ \hline\end{array}\]](http://latex.artofproblemsolving.com/a/f/f/aff8c0c2bd3960b9c54ff61ff338bd697fa84bcd.png)
Solution
A simple method of solving the problem is trying each of the answer choices.
One can plug in values of x and y for each, because many x-values and their single corresponding y-values are given.
Choice A works for (1,3) and (2,7) but fails to work on (3,13) because 3*4=12 and 12-1=11, not 13.
Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7.
Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31.
Thus, the answer is 
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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. 
