2010 AMC 8 Problem 25

Attempt Problem 25 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Every day at school, Jo climbs a flight of 66 stairs. Jo can take the stairs 1,1, 2,2, or 33 at a time. For example, Jo could climb 3,3, then 1,1, then 2.2. In how many ways can Jo climb the stairs?
 13 \ 13
 18 \ 18
 20 \ 20
 22 \ 22
 24 \ 24
Answer: E
Concepts:partitions and compositionsrecursive counting
Small hint:
The first step can be 11, 22, or 33 stairs, giving a recurrence.
Big hint:
Let wnw_n be the number of ways to climb nn stairs.
Solution:
Let wnw_n be the number of ways to climb nn stairs. For n≥4n\ge4, the first step can be 11, 22, or 33 stairs, so wn=wn−1+wn−2+wn−3w_n=w_{n-1}+w_{n-2}+w_{n-3}.
We have w1=1w_1=1, w2=2w_2=2, and w3=4w_3=4. Therefore w4=4+2+1=7w_4=4+2+1=7, w5=7+4+2=13w_5=7+4+2=13, and w6=13+7+4=24w_6=13+7+4=24.
Thus, the answer is E .
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