Difference between revisions of "2020 AMC 10B Problems/Problem 15"
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Similarly, if we start with <math>b_1, b_2, b_3, \cdots,</math> and remove all multiples of <math>4</math>, and get <math>c_1, c_2, \cdots</math> | Similarly, if we start with <math>b_1, b_2, b_3, \cdots,</math> and remove all multiples of <math>4</math>, and get <math>c_1, c_2, \cdots</math> | ||
− | We have <math>c_{3n+ | + | We have <math>c_{3n+1}=b_{4n+1},c_{3n+2}=b_{4n+2}, c_{3n+3}=b_{4n+3}</math>. |
If we start with <math>c_1, c_2, \cdots</math> and <math>d_1, d_2, \cdots </math> are remove all multiples of <math>5</math>, we get <cmath>d_{4n+1}=c_{5n+1}, d_{4n+2}=c_{5n+2}, d_{4n+3}=c_{5n+3}, d_{4n+4}=c_{5n+4}.</cmath> | If we start with <math>c_1, c_2, \cdots</math> and <math>d_1, d_2, \cdots </math> are remove all multiples of <math>5</math>, we get <cmath>d_{4n+1}=c_{5n+1}, d_{4n+2}=c_{5n+2}, d_{4n+3}=c_{5n+3}, d_{4n+4}=c_{5n+4}.</cmath> | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
and the answer is <math>4+2+5=\boxed{\textbf{(D)}\ 11}</math>. | and the answer is <math>4+2+5=\boxed{\textbf{(D)}\ 11}</math>. | ||
+ | |||
+ | ~szhangmath | ||
==Video Solution (HOW TO CRITICALLY THINK!!!)== | ==Video Solution (HOW TO CRITICALLY THINK!!!)== |
Latest revision as of 13:57, 30 October 2023
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
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
Video Solution
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 |
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