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 123451234512β¦. He then erased every third digit from his list (that is, the 3rd, 6th, 9th, ... digits from the left), then erased every fourth digit from the resulting list (that is, the 4th,8th,12th,β¦ 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 positions 2019,2020, and 2021?
Answer Choices:
A. 7
B. 9
C. 10
D. 11
E. 12
Solution:
Note that cycles exist initially and after each round of erasing. Let the parentheses denote cycles. It follows that: 0. Initially, the list has cycles of length 5:
(12345)=12345123451234512345β―
1. To find one cycle after the first round of erasing, we need one cycle of length lcm(3,5)=15 before erasing. So, we first group 515β=3 copies of the current cycle into one, then erase:
(12345)ββΆ(123451234512345)βΆ(124523451234)βΆ(1245235134)β
As a quick confirmation, one cycle should have length 15β
(1β31β)=10 at this point. 2. To find one cycle after the second round of erasing, we need one cycle of length lcm(4,10)=20 before erasing. So, we first group 1020β=2 copies of the current cycle into one, then erase:
(1245235134)ββΆ(12452351341245235134)βΆ(124ξ 235ξ 341ξ 2452ξ Β²134) βΆ(124235341452513).β
As a quick confirmation, one cycle should have length 20β
(1β41β)=15 at this point. 3. To find one cycle after the third round of erasing, we need one cycle of length lcm(5,15)=15 before erasing. We already have it here, so we erase:
(124235341452513)ββΆ(124225341A5251))βΆ(124253415251).β
As a quick confirmation, one cycle should have length 15β
(1β51β)=12 at this point.
Since 2019,2020,2021 are congruent to 3,4,5 modulo 12, respectively, the three digits in the final positions 2019,2020,2021 are 4,2,5, respectively.
Therefore, the answer is 4+2+5= (D)11β.
OR
β After erasing every third digit, the list becomes 1245235134 . . . repeated.
β After erasing every fourth digit from this list, the list becomes 124235341452513β¦ repeated.
β Finally, after erasing every fifth digit from this list, the list becomes 124253415251β¦repeated.
Since this list repeats every 12 digits and 2019,2020,2021 are 3,4,5 respectively in (mod12), we have that the 2019th, 2020th, and 2021st digits are the 3rd, 4th, and 5th digits respectively. It follows that the answer is 4+2+5=(D)11β.
The problems on this page are the property of the MAA's American Mathematics Competitions