Difference between revisions of "2020 AMC 10B Problems/Problem 15"
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As the LCM of 3, 4, and 5 is <math>60</math>, let's look at a 60-digit block of original numbers (many will be erased by Steve). After he erases every third number (1/3), then every fourth number of what remains (1/4), then every fifth number of what remains (1/5), we are left with <math>\dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot 60=24</math> digits from this 60-digit block. <math>2019 \equiv 3 \pmod {24}, 2020 \equiv 4 \pmod {24}, 2021 \equiv 5 \pmod {24}</math>. Writing out the first few digits of this sequence, we have <math>\underbrace{1}_{\#1}, \underbrace{2}_{\#2}, \cancel{3}, \underbrace{4}_{\#3}, \cancel{5}, \cancel{1}, \underbrace{2}_{\#4}, \cancel{3}, \cancel{4}, \underbrace{5}_{\#5}, \dots</math>. Therefore, our answer is <math>4+2+5=\boxed{\textbf{(D) }11}</math>. | As the LCM of 3, 4, and 5 is <math>60</math>, let's look at a 60-digit block of original numbers (many will be erased by Steve). After he erases every third number (1/3), then every fourth number of what remains (1/4), then every fifth number of what remains (1/5), we are left with <math>\dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot 60=24</math> digits from this 60-digit block. <math>2019 \equiv 3 \pmod {24}, 2020 \equiv 4 \pmod {24}, 2021 \equiv 5 \pmod {24}</math>. Writing out the first few digits of this sequence, we have <math>\underbrace{1}_{\#1}, \underbrace{2}_{\#2}, \cancel{3}, \underbrace{4}_{\#3}, \cancel{5}, \cancel{1}, \underbrace{2}_{\#4}, \cancel{3}, \cancel{4}, \underbrace{5}_{\#5}, \dots</math>. Therefore, our answer is <math>4+2+5=\boxed{\textbf{(D) }11}</math>. | ||
~BakedPotato66 | ~BakedPotato66 | ||
| + | |||
| + | ==Solution 3 (Illustrations)== | ||
| + | <b>I will finish shortly. No edit please.</b> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 02:15, 16 April 2021
Contents
Problem
Steve wrote the digits 
, 
, 
, 
, and 
in order repeatedly from left to right, forming a list of 
digits, beginning 
He then erased every third digit from his list (that is, the rd, 
th, 
th, 
digits from the left), then erased every fourth digit from the resulting list (that is, the th, 
th, 
th, 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 
?
Solution 1
After erasing every third digit, the list becomes 
repeated. After erasing every fourth digit from this list, the list becomes 
repeated. Finally, after erasing every fifth digit from this list, the list becomes 
repeated. Since this list repeats every digits and since 
are 
respectively
in 
we have that the 
th, 
th, and 
st digits are the rd, th, and th digits respectively. It follows that the answer is 
Solution 2
As the LCM of 3, 4, and 5 is 
, let's look at a 60-digit block of original numbers (many will be erased by Steve). After he erases every third number (1/3), then every fourth number of what remains (1/4), then every fifth number of what remains (1/5), we are left with 
digits from this 60-digit block. 
. Writing out the first few digits of this sequence, we have 
. Therefore, our answer is 
.
~BakedPotato66
Solution 3 (Illustrations)
I will finish shortly. No edit please.
~MRENTHUSIASM
Video Solution
https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
