2020 AMC 10B Problems/Problem 15
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 of Solutions 1 and 2)
Note that cycles exist initially and after each round of erasing.
Two solutions follow from here:
Solution 3.1 (Stepwise: Similar to Solution 1)
Let the parentheses denote cycles. Initially, the list has cycles of length
- To find one cycle after the first round of erasing, we need one cycle of length
before erasing. So, we first group
copies of the current cycle into one, then erase:
As a quick verification, one cycle should have length
at this point.
- To find one cycle after the second round of erasing, we need one cycle of length
before erasing. So, we first group
copies of the current cycle into one, then erase:
As a quick verification, one cycle should have length
at this point.
- To find one cycle after the third round of erasing, we need one cycle of length
before erasing. We already have it here, so we erase:
As a quick verification, one cycle should have length
at this point.
As are congruent to
modulo
respectively, the three digits in the final positions
are
respectively:
The answer is
~MRENTHUSIASM (inspired by TheBeautyofMath)
Solution 3.2 (Direct: Similar to Solution 2)
We will consider one cycle after all three rounds of erasing.
Suppose this cycle has length before any round of erasing. It follows that
- After the first round of erasing, one cycle will have length
from which
must be divisible by
- After the second round of erasing, one cycle will have length
from which
must be divisible by
- After the third round of erasing, one cycle will have length
from which
must be divisible by
The least such positive integer is
Therefore, the pattern will repeat for every
digits in the original list.
Four groups of digits in the original list are shown below. The digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively:
Diagram in progress
As confirmed, each group has digits remaining.
Since are congruent to
modulo
respectively, the three digits in the final positions
are
respectively:
Diagram in progress
The answer is
~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 | |
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