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Jacob's Complicated Notation

Source: AMC 10 2008A Problem 22

February 22, 2008
probabilityAMC

Problem Statement

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1 1. If it comes up tails, he takes half of the previous term and subtracts 1 1. What is the probability that the fourth term in Jacob's sequence is an integer? <spanclass=latexbold>(A)</span> 16<spanclass=latexbold>(B)</span> 13<spanclass=latexbold>(C)</span> 12<spanclass=latexbold>(D)</span> 58<spanclass=latexbold>(E)</span> 34 <span class='latex-bold'>(A)</span>\ \frac{1}{6} \qquad <span class='latex-bold'>(B)</span>\ \frac{1}{3} \qquad <span class='latex-bold'>(C)</span>\ \frac{1}{2} \qquad <span class='latex-bold'>(D)</span>\ \frac{5}{8} \qquad <span class='latex-bold'>(E)</span>\ \frac{3}{4}