Difference between revisions of "2022 AMC 12B Problems/Problem 3"
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Let <math>P(a,b)</math> denote the digit <math>a</math> written <math>b</math> times and let <math>\overline{a_1a_2\cdots a_n}</math> denote the concatenation of <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math>. | Let <math>P(a,b)</math> denote the digit <math>a</math> written <math>b</math> times and let <math>\overline{a_1a_2\cdots a_n}</math> denote the concatenation of <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math>. | ||
− | Observe that <cmath>\overline{P(1,n) 2 P(1,n)} = \overline{P(1,n+1)P(0,n)} + P(1,n+1).</cmath> | + | Observe that <cmath>\overline{P(1,n) \: 2 \: P(1,n)} = \overline{P(1,n+1) \: P(0,n)} + P(1,n+1).</cmath> |
− | Since <math>\overline{k P(0,n)} = k \cdot 10^n</math> for all positive integers <math>k</math> and <math>n</math>, <math>\overline{P(1,n+1)P(0,n)} + P(1,n+1)</math> is equal to | + | Since <math>\overline{k \: P(0,n)} = k \cdot 10^n</math> for all positive integers <math>k</math> and <math>n</math>, <math>\overline{P(1,n+1) \: P(0,n)} + P(1,n+1)</math> is equal to |
<cmath>P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).</cmath> | <cmath>P(1,n+1) \cdot 10^n + P(1,n+1) = (P(1,n+1))(10^n + 1).</cmath> | ||
Revision as of 06:55, 18 November 2022
Contents
Problem
How many of the first ten numbers of the sequence , , , ... are prime numbers?
Solution 1
Write , , . It becomes clear that of these numbers are prime.
In general, (where there are 's on either side of the ) can be written as , where the first term has 's.
Solution 2
Let denote the digit written times and let denote the concatenation of , , ..., .
Observe that
Since for all positive integers and , is equal to
Both terms are integers larger than since , so of the numbers of the sequence are prime.
See Also
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.