Difference between revisions of "2005 Alabama ARML TST Problems/Problem 10"
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==Solution== | ==Solution== | ||
+ | === Solution 1 === | ||
He can either take 14 one-steps, or 12 one-steps and 1 two-step, etc., so we have | He can either take 14 one-steps, or 12 one-steps and 1 two-step, etc., so we have | ||
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=1+13+66+165+210+126+28+1=14+231+336+29=245+365=\boxed{610} | =1+13+66+165+210+126+28+1=14+231+336+29=245+365=\boxed{610} | ||
\end{eqnarray}</math> | \end{eqnarray}</math> | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>a_n</math> represent the number of sequences of steps that can be taken up the stairs. Clearly <math>a_{1} = 1, a_{2} = 2</math>. Any sequence of length <math>n + 1</math> is either composed of a sequence of length <math>n</math> followed by another step or a sequence of length <math>n-1</math> followed by two steps. Thus we have the recursion | ||
+ | \[ | ||
+ | a_{n + 1} = a_{n} + a_{n - 1} | ||
+ | \] | ||
+ | which indeed is the Fibonacci sequence, except with indexes shifted. Matching them, we have <math>a_n = F_{n + 1}</math>, and <math>a_{14} = F_{15} = \boxed{610}</math>. | ||
==See also== | ==See also== | ||
{{ARML box|year=2005|state=Alabama|num-b=9|num-a=11}} | {{ARML box|year=2005|state=Alabama|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Revision as of 17:21, 1 March 2008
Contents
[hide]Problem
When Jon Stewart walks up stairs he takes one or two steps at a time. His stepping sequence is not necessarily regular. He might step up one step, then two, then two again, then one, then one, and then two in order to climb up a total of 9 steps. In how many ways can Jon walk up a 14 step stairwell?
Solution
Solution 1
He can either take 14 one-steps, or 12 one-steps and 1 two-step, etc., so we have
$
Solution 2
Let represent the number of sequences of steps that can be taken up the stairs. Clearly
. Any sequence of length
is either composed of a sequence of length
followed by another step or a sequence of length
followed by two steps. Thus we have the recursion
\[
a_{n + 1} = a_{n} + a_{n - 1}
\]
which indeed is the Fibonacci sequence, except with indexes shifted. Matching them, we have
, and
.
See also
2005 Alabama ARML TST (Problems) | ||
Preceded by: Problem 9 |
Followed by: Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |