Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,4,6,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,4,6,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> | ||
(AOPS12142015) | (AOPS12142015) | ||
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| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Revision as of 14:05, 16 February 2018
Problem
How many of the first 
numbers in the sequence 
are divisible by 
?
Solution
Note that 
for some odd 
will suffice 
. Each 
, so the answer is 
(AOPS12142015)
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
