1952 AHSME Problems/Problem 40

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Problem

In order to draw a graph of $ax^2+bx+c$, a table of values was constructed. These values of the function for a set of equally spaced increasing values of $x$ were $3844, 3969, 4096, 4227, 4356, 4489, 4624$, and $4761$. The one which is incorrect is:

$\text{(A) } 4096 \qquad \text{(B) } 4356 \qquad \text{(C) } 4489 \qquad \text{(D) } 4761 \qquad \text{(E) } \text{none of these}$


Solution

Since the polynomial is quadratic, its second differences must be constant. Taking the first differences, we have \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*} This leads to a common second difference of $2$, with the only discrepancy around the point $4227$. Observe that if this point were instead $4225$, the common second difference would, indeed be $2$ for all data points. Therefore the answer is $4227$, or $\fbox{E}$

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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