Difference between revisions of "1988 AHSME Problems/Problem 30"
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<math>\textbf{(A)}\ \text{0}\qquad | <math>\textbf{(A)}\ \text{0}\qquad | ||
\textbf{(B)}\ \text{1 or 2}\qquad | \textbf{(B)}\ \text{1 or 2}\qquad | ||
| − | \textbf{(C)}\ \text{3, 4, 5 or 6}\qquad | + | \textbf{(C)}\ \text{3, 4, 5 or 6}\qquad |
| − | \textbf{(D)}\ \text{more than 6 but finitely many}\qquad | + | \textbf{(D)}\ \text{more than 6 but finitely many}\qquad |
| − | \textbf{(E) }\infty</math> | + | \textbf{(E) }\infty</math> |
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==Solution== | ==Solution== | ||
| − | + | Note that <math>x_{0} = 0</math> gives the constant sequence <math>0, 0, ...</math>, since <math>f(0) = 4 \cdot 0 - 0^2 = 0</math>. Because <math>f(4)=0, x_{0} = 4</math> gives the sequence <math>4, 0, 0, ...</math> with two different values. Similarly, <math>f(2) = 4</math>, so <math>x_{0} = 2</math> gives the sequence <math>2, 4, 0, 0, ...</math> with three values. In general, if <math>x_{0} = a_{n}</math> gives the sequence <math>a_{n}, a_{n-1}, ... , a_{2}, a_{1}, ...</math> with <math>n</math> different values, and <math>f(a_{n+1}) = a_{n}</math>, then <math>x_{0} = a_{n+1}</math> gives a sequence with <math>n+1</math> different values. (It is easy to see that we could not have <math>a_{n+1} = a_{i}</math> for some <math>i < n + 1</math>.) Thus, it follows by induction that there is a sequence with <math>n</math> distinct values for every positive integer <math>n</math>, as long as we can verify that there is always a real number <math>a_{n+1}</math> such that <math>f(a_{n+1}) = a_{n}</math>. This makes the answer <math>\boxed{\text{E}}</math>. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to <math>f(a_{n+1}) = 4a_{n+1} - a_{n+1}^{2} = a_{n}</math> are <math>a_{n+1} = 2 \pm \sqrt{4 - a_{n}}</math>. Hence if <math>0 \leq a_{n} \leq 4</math>, then <math>a_{n+1}</math> is real, since the part under the square root is non-negative, and in fact <math>0 \leq a_{n+1} \leq 4</math>, since <math>4-a_{n}</math> will be between <math>0</math> and <math>4</math>, so the square root will be between <math>0</math> and <math>2</math>, and <math>2 \pm</math> something between <math>0</math> and <math>2</math> gives something between <math>0</math> and <math>4</math>. Finally, since <math>a_{1} = 0 \leq 4</math>, it follows by induction that all terms satisfy <math>0 \leq a_{n} \leq 4</math>; in particular, they are all real. | |
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== See also == | == See also == | ||
Latest revision as of 15:16, 27 February 2018
Problem
Let 
. Give 
, consider the sequence defined by 
for all 
.
For how many real numbers will the sequence 
take on only a finite number of different values?
Solution
Note that 
gives the constant sequence 
, since 
. Because 
gives the sequence 
with two different values. Similarly, 
, so 
gives the sequence 
with three values. In general, if 
gives the sequence 
with 
different values, and 
, then 
gives a sequence with 
different values. (It is easy to see that we could not have 
for some 
.) Thus, it follows by induction that there is a sequence with distinct values for every positive integer , as long as we can verify that there is always a real number 
such that . This makes the answer 
. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to 
are 
. Hence if 
, then is real, since the part under the square root is non-negative, and in fact 
, since 
will be between 
and 
, so the square root will be between and 
, and 
something between and gives something between and . Finally, since 
, it follows by induction that all terms satisfy ; in particular, they are all real.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Question | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
