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 (Simulation)
Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:
- 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 confirmation, 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 confirmation, 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 confirmation, one cycle should have length at this point.
Since are congruent to modulo respectively, the three digits in the final positions are respectively: Therefore, the answer is
~MRENTHUSIASM (inspired by TheBeautyofMath)
Solution 2 (Simulation)
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 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 3 (Analysis)
Note that cycles exist initially and after each round of erasing.
We will consider one cycle after all three rounds of erasing. Suppose this cycle has length before any round of erasing. It follows that:
- Initially, one cycle has length from which must be divisible by
- After the first round of erasing, one cycle has length from which must be divisible by
- After the second round of erasing, one cycle has length from which must be divisible by
- After the third round of erasing, one cycle has length from which must be divisible by
The least such positive integer is So, there is a repeating pattern for every digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: As indicated by the white squares, each group of digits on the original list has digits remaining on the final list.
Since are congruent to modulo respectively, the three digits in the final positions are respectively: Therefore, the answer is
~MRENTHUSIASM
Solution 4 (Analysis)
As the LCM of , , and is , let us look at a -digit block of original numbers (many will be erased by Steve). After he erases every third number , then every fourth number of what remains , then every fifth number of what remains , we are left with digits from this -digit block. . Writing out the first few digits of this sequence, we have . Therefore, our answer is .
~BakedPotato66
Solution 5
Lemma: Given a sequence , and an positive integer . If we erase every th item in this sequence, and we name as the remaining sequence. Then we have
Proof: For with some , we will have items removed before this item, so it becomes in the new sequence. Hence, we have .
If we start with and let be the sequence after removing all the indexes that are multiples of . Then, we have .
Similarly, if we start with and remove all multiples of , and get We have .
If we start with and are remove all multiples of , we get Therefore, and the answer is .
~szhangmath
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See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
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