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Difference between revisions of "2016 AMC 12B Problems/Problem 7"

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==Solution==
 
==Solution==
 
Following the pattern, you are crossing out...
 
Following the pattern, you are crossing out...
−
Time 1: Every non-multiple of 2
+
 
−
Time 2: Every non-multiple of 4
+
Time 1: Every non-multiple of <math>2</math>
−
Time 3: Every non-multiple of 8
+
 
−
Following this pattern, you are left with every non-multiple of 64 which is only 64.
+
Time 2: Every non-multiple of <math>4</math>
 +
 
 +
Time 3: Every non-multiple of <math>8</math>
 +
 
 +
Following this pattern, you are left with every multiple of <math>64</math> which is only <math>\boxed{\textbf{(D)}64}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2016|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:38, 12 August 2018

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

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 $\boxed{\textbf{(D)}64}$.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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