Difference between revisions of "2018 AMC 10B Problems/Problem 13"
Aops12142015 (talk | contribs) (→Solution) |
Brainpopper (talk | contribs) |
||
| Line 13: | Line 13: | ||
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) | ||
| + | |||
| + | ==Solution 2== | ||
| + | If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots. We divide </math>2018<math> by </math>4<math> to get </math>504<math> with </math>2<math> left over. One divisible number will be in the </math>2<math> left over, so out answer is </math>\boxed{\textbf{(C) } 505}$. | ||
==See Also== | ==See Also== | ||
Revision as of 09:32, 17 February 2018
Contents
Problem
How many of the first 
numbers in the sequence 
are divisible by 
?
Solution
Note that 
for some odd 
will suffice 
. Each 
, so the answer is 
(AOPS12142015)
Solution 2
If we divide each number by , we see a pattern occuring in every 4 numbers. 
2018
4
504
2
2
\boxed{\textbf{(C) } 505}$.
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. 
