2020 AMC 10B Problems/Problem 15: Difference between revisions

Revision as of 20:33, 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 $2019^{th}$ , $2020^{th}$ , and the $2021^{st}$ numbers are $2, 4, 5$ , giving the answer $2+4+5= \boxed {\textbf{(D)} \text{ 11}}$ ~DragonWarrior123

Solution 2

After erasing every third digit, the list becomes $1245235134\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\ldots$ repeated. Since this list repeats every $12$ digits and since $2019,2020,2020$ are $3,4,5$ respectively in $\pmod{12},$ we have that the $2019$ th, $2020$ th, and $2021$ st digits are the $3$ rd, $4$ th, and $5$ th digits respectively. It follows that the answer is $2+4+5= \boxed {\textbf{(D)} \text{ 11}}.$ ~dolphin7

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

@@ Line 10: / Line 10: @@
 ==Solution 2==
-After erasing every third digit, the list becomes <math>1245235134\ldots</math> repeated. After erasing every fourth digit, the list becomes <math>124235341452513\ldots</math> repeated. Finally, after erasing every fourth digit, the list becomes <math>124253415251\ldots</math> repeated. Since this list repeats every <math>12</math> digits and since <math>2019,2020,2020</math> are <math>3,4,5</math> respectively in <math>\pmod{12},</math> we have that the <math>2019</math>th, <math>2020</math>th, and <math>2021</math>st digits are the <math>3</math>rd, <math>4</math>th, and <math>5</math>th digits respectively. It follows that the answer is <math>2+4+5= \boxed {\textbf{(D)} \text{ 11}}.</math>
+After erasing every third digit, the list becomes <math>1245235134\ldots</math> repeated. After erasing every fourth digit from this list, the list becomes <math>124235341452513\ldots</math> repeated. Finally, after erasing every fifth digit from this list, the list becomes <math>124253415251\ldots</math> repeated. Since this list repeats every <math>12</math> digits and since <math>2019,2020,2020</math> are <math>3,4,5</math> respectively in <math>\pmod{12},</math> we have that the <math>2019</math>th, <math>2020</math>th, and <math>2021</math>st digits are the <math>3</math>rd, <math>4</math>th, and <math>5</math>th digits respectively. It follows that the answer is <math>2+4+5= \boxed {\textbf{(D)} \text{ 11}}.</math>
 ~dolphin7

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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