SAT1000 - P500

Algebra Level pending

{ a n } \{a_n\} is a sequence such that a 1 = m ( m ∈ N + ) a_1=m\ (m \in \mathbb N^+) , a n + 1 = { a n 2 , a n ≡ 0 ( m o d 2 ) 3 a n + 1 , a n ≡ 1 ( m o d 2 ) a_{n+1}=\begin{cases} \dfrac{a_n}{2} ,\ a_n \equiv 0 \pmod{2} \\ 3 a_n+1 ,\ a_n \equiv 1 \pmod{2} \end{cases} If a 6 = 1 a_6=1 , find the sum of all possible value(s) for m m .

Have a look at my problem set: SAT 1000 problems


The answer is 41.

Alice Smith by Alice Smith , 20, China

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1 solution

Ved Pradhan
Jun 18, 2020

Work backwards! Given a number a n a_n , a n − 1 a_{n-1} can be 2 a n 2a_n or a n − 1 3 \dfrac{a_{n}-1}{3} , if it's natural. Thus, going backwards, we get this:

1 → 2 → 4 → { 1 → 2 → 4 8 → 16 → { 5 32 1\rightarrow 2\rightarrow 4\rightarrow \begin{cases} 1\rightarrow 2\rightarrow 4 \\ 8\rightarrow 16\rightarrow \begin{cases} 5 \\ 32 \end{cases} \end{cases}

We have three possibilities for a 1 a_1 . Their sum is 4 + 5 + 32 = 41 4+5+32=\boxed{41} .

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