AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 21 Hard

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1. If it comes up tails, he takes half of the previous term and subtracts

1. The probability that the n-th term is an integer is less than \frac{1}{2}. What is the minimum value of n?

  • A.

    3

  • B.

    4

  • C.

    5

  • D.

    6

  • E.

    7

Answer:D

We construct a tree showing all possible outcomes that Jacob may get after 3 flips; we can do this because there are only 8 possibilities:6\left\{ \begin{array}{l} {H:11\left\{ \begin{array}{l} {H:21\left\{ \begin{array}{l} {H:\boxed {41}}\\{T:9.5} \end{array}\right. }\\{T:4.5\left\{ \begin{array}{l} {H:\boxed {8}}\\{T:1.25} \end{array}\right. } \end{array}\right. }\\{T:2\left\{ \begin{array}{l} {H:3\left\{ \begin{array}{l} {H:\boxed {5}}\\{T:0.5} \end{array}\right. }\\{T:0\left\{ \begin{array}{l} {H:\boxed {-1}}\\{T:\boxed {-1}} \end{array}\right. } \end{array}\right. } \end{array}\right.

Similarily, we found the probabililty for the 5-th term to be an integer is \frac{1}{2}, for the 6-th term to be an integer is \frac{15}{32}. So the answer is 6.

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