Difference between revisions of "2016 AMC 12B Problems/Problem 7"
(Created page with "==Problem== Josh writes the numbers <math>1,2,3,\dots,99,100</math>. He marks out <math>1</math>, skips the next number <math>(2)</math>, marks out <math>3</math>, and contin...") |
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==Solution== | ==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 non-multiple of 64 which is only 64. | ||
==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}} | ||
Revision as of 13:37, 21 February 2016
Problem
Josh writes the numbers 
. He marks out 
, skips the next number 
, marks out 
, 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 , skips the next number 
, marks out 
, skips 
, marks out 
, and so on to the end. Josh continues in this manner until only one number remains. What is that number?
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 non-multiple of 64 which is only 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
