2016 AMC 12B Problems/Problem 7

Problem

Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?

$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$

Solution 1

Following the pattern, you are crossing out...

Time 1: Every non-multiple of $2$

Time 2: Every non-multiple of $4$

Time 3: Every non-multiple of $8$

Following this pattern, you are left with every multiple of $64$ which is only $\boxed{\textbf{(D)}64}$ .

Solution 2 (Blitz)

$\boxed{\textbf{(A)}13}$ is odd so it is automatically crossed out in the first selection.

$\boxed{\textbf{(B)}32}$ and $\boxed{\textbf{(E)}96}$ behave similarly so they would both get knocked out in the same round.

$\boxed{\textbf{(C)}56}$ and $\boxed{\textbf{(D)}64}$ are left, and since it is (slightly) more likely for the number to be larger, we can guess $\boxed{\textbf{(D)}64}$ if we're really running out of time.

~panikd

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2016 AMC 12B Problems/Problem 7

Contents

Problem

Solution 1

Solution 2 (Blitz)

See Also