Difference between revisions of "1950 AHSME Problems/Problem 16"

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== Problem==
 
== Problem==
  
The formula which expresses the relationship between <math>x</math> and <math>y</math> as shown in the accompanying table is:
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The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:
  
<cmath> \begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\hline y&100&90&70&40&0\\hline\end{tabular} </cmath>
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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
  
<math> \textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these} </math>
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==Solution==
  
==Solution==
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Use properties of exponents to move the squares outside the brackets use difference of squares.
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<cmath>[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4</cmath>
  
Plug in the points <math>(0,100)</math> and <math>(4,0)</math> into each equation. The only one that works for both points is <math>\mathrm{(C)}.</math> Plug in the rest of the points to confirm the answer is indeed <math>\boxed{\mathrm{(C)}\ y=100-5x-5x^2.}</math>
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Using the binomial theorem, we can see that the number of terms is <math>\boxed{\mathrm{(B)}\ 5.}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1950|num-b=15|num-a=17}}
 
{{AHSME box|year=1950|num-b=15|num-a=17}}

Revision as of 21:18, 13 November 2011

Problem

The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Use properties of exponents to move the squares outside the brackets use difference of squares.

\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]

Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5.}$

See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AHSME Problems and Solutions