Difference between revisions of "2022 AMC 10B Problems/Problem 6"
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+ | {{duplicate|[[2022 AMC 10B Problems/Problem 6|2022 AMC 10B #6]] and [[2022 AMC 12B Problems/Problem 3|2022 AMC 12B #3]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
How many of the first ten numbers of the sequence <math>121, 11211, 1112111, \ldots</math> are prime numbers? | How many of the first ten numbers of the sequence <math>121, 11211, 1112111, \ldots</math> are prime numbers? | ||
Line 6: | Line 8: | ||
==Solution 1 (Generalization)== | ==Solution 1 (Generalization)== | ||
The <math>n</math>th term of this sequence is | The <math>n</math>th term of this sequence is | ||
− | <cmath>\sum_{k=n}^{2n} | + | <cmath>\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.</cmath> |
It follows that the terms are | It follows that the terms are | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Line 18: | Line 20: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2 (Educated | + | ==Solution 2 (Detailed Explanation of Solution 1)== |
+ | |||
+ | Denote this sequence as <math>a_{n}</math>, then we can find that | ||
+ | <cmath>\begin{align*} | ||
+ | a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ | ||
+ | a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ | ||
+ | a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ | ||
+ | & \ \vdots | ||
+ | \end{align*}</cmath> | ||
+ | So, we can induct that the general term is | ||
+ | <cmath>\begin{align*} | ||
+ | a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ | ||
+ | &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ | ||
+ | &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. | ||
+ | \end{align*}</cmath> | ||
+ | Therefore, there are <math>\boxed{\textbf{(A) } 0}</math> prime numbers in this sequence. | ||
+ | |||
+ | ~PythZhou | ||
+ | |||
+ | ==Solution 3 (Simple Sums)== | ||
+ | |||
+ | Observe how | ||
+ | <cmath>\begin{align*} | ||
+ | 121 &= 110 + 11, \\ | ||
+ | 11211 &= 11100 + 111, \\ | ||
+ | 1112111 &= 1111000 + 1111, \\ | ||
+ | & \ \vdots | ||
+ | \end{align*}</cmath> | ||
+ | all take the form of <cmath>\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).</cmath> | ||
+ | Factoring each of the sums, we have <cmath>11(10+1), 111(100+1), 1111(1000+1), \ldots</cmath> respectively. With each number factored, there are <math>\boxed{\textbf{(A) } 0}</math> primes in the set. | ||
+ | |||
+ | ~ab2024 | ||
+ | |||
+ | ==Solution 4 (Educated Guess)== | ||
− | Note | + | Note that <math>121</math> is divisible by <math>11</math> and <math>11211</math> is divisible by <math>3</math>. Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is <math>\boxed{\textbf{(A) } 0}.</math> |
~Dhillonr25 | ~Dhillonr25 | ||
+ | |||
+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
+ | |||
+ | ~~Hayabusa1 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/_KNR0JV5rdI?t=562 | ||
+ | |||
+ | ==Video Solution by Math4All999== | ||
+ | https://youtu.be/5QYh3hNaDa0?feature=shared | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}} | ||
+ | {{AMC12 box|year=2022|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:18, 20 March 2024
- The following problem is from both the 2022 AMC 10B #6 and 2022 AMC 12B #3, so both problems redirect to this page.
Contents
Problem
How many of the first ten numbers of the sequence are prime numbers?
Solution 1 (Generalization)
The th term of this sequence is
It follows that the terms are
Therefore, there are
prime numbers in this sequence.
~MRENTHUSIASM
Solution 2 (Detailed Explanation of Solution 1)
Denote this sequence as , then we can find that
So, we can induct that the general term is
Therefore, there are
prime numbers in this sequence.
~PythZhou
Solution 3 (Simple Sums)
Observe how
all take the form of
Factoring each of the sums, we have
respectively. With each number factored, there are
primes in the set.
~ab2024
Solution 4 (Educated Guess)
Note that is divisible by
and
is divisible by
. Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is
~Dhillonr25
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=562
Video Solution by Math4All999
https://youtu.be/5QYh3hNaDa0?feature=shared
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.