Difference between revisions of "1952 AHSME Problems/Problem 40"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | Since the polynomial is quadratic, its second differences must be constant. Taking the first differences, we have |
| + | <cmath>\begin{align*} | ||
| + | 3969-3844&=125 \\ | ||
| + | 4096-3969&=127 \\ | ||
| + | 4227-4096&=131 \\ | ||
| + | 4356-4227&=129 \\ | ||
| + | 4489-4356&=133 \\ | ||
| + | 4624-4489&=135 \\ | ||
| + | 4761-4624&=137 | ||
| + | \end{align*}</cmath> | ||
| + | This leads to a common second difference of <math>2</math>, with the only discrepancy around the point <math>4227</math>. Observe that if this point were instead <math>4225</math>, the common second difference would, indeed be <math>2</math> for all data points. Therefore the answer is <math>4227</math>, or <math>\fbox{E}</math> | ||
== See also == | == See also == | ||
Latest revision as of 18:34, 28 April 2020
Problem
In order to draw a graph of 
, a table of values was constructed. These values of the function for a set of equally spaced increasing values of 
were 
, and 
. The one which is incorrect is:
Solution
Since the polynomial is quadratic, its second differences must be constant. Taking the first differences, we have
This leads to a common second difference of 
, with the only discrepancy around the point 
. Observe that if this point were instead 
, the common second difference would, indeed be for all data points. Therefore the answer is , or 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
Followed by Problem 41 | |
| 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. 
