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

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(Solution)
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==Solution==
 
==Solution==
Bash it out until you find a pattern.
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Bash it out until you find a pattern. In the end, you'll see that the <math>2018^{th}</math>, <math>2019^{th}</math>, <math>2020^{th}</math> are <math>2, 4, 5</math>, giving the answer <math>2+4+5= \boxed {D)11}</math>
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~DragonWarrior123
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:45, 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

Bash it out until you find a pattern. In the end, you'll see that the $2018^{th}$, $2019^{th}$, $2020^{th}$ are $2, 4, 5$, giving the answer $2+4+5= \boxed {D)11}$ ~DragonWarrior123

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

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