Question

Josh writes the numbers $1,2,3,\ldots,99,100.$ 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 his 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?

$13$

$32$

$56$

$64$

$96$

Solution:

The first pass removes the odd numbers, leaving the multiples of $2.$ The second pass removes $2,6,10,\ldots,$ leaving the multiples of $4.$ In general, after the $n$ th pass only the multiples of $2^n$ remain. The surviving number is the highest power of $2$ not exceeding $100,$ which is $2^6=64.$

Thus, the correct answer is D.

2016 AMC 12B Problem 7

Solution:

Problem 7 in Other Years