Difference between revisions of "2020 AMC 10B Problems/Problem 15"

m
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
 +
 +
==Video Solution==
 +
https://youtu.be/t6yjfKXpwDs
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2020|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2020|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:35, 7 February 2020

Problem

Steve wrote the digits $1$, $2$, $3$, $4$, and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$rd, $6$th, $9$th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$th, $8$th, $12$th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$?

$\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}$

Solution

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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. AMC logo.png