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1966 AHSME Problems/Problem 7

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Problem

Let \frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2} be an identity in x. The numerical value of N_1N_2 is:

\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246

Solution

\frac {N_1}{x - 1} + \frac {N_2}{x - 2} = \frac{N_1(x-2) + N_2(x-1)}{(x-1)(x-2)} = \frac{(N_1 + N_2)x - (2N_1 + N_2)}{x^2 - 3...

Comparing coefficients, we have the system of two equations

\begin{align*}N_1 + N_2 &= 35\\N_1 + 2N_2 &= 29\end{align*}

Subtracting the first equation from the second yields N_2 = -6. Substituting yields N_1 = 41, and their product is -246 \Rightarrow \mathrm{(A)}.

See also

1966 AHSME (Problems)
Preceded by
Problem 6
Followed by
Problem 8
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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