Difference between revisions of "2022 AMC 12A Problems/Problem 24"
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than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it | than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it | ||
does not contain at least <math>2</math> digits less than <math>2</math>.) | does not contain at least <math>2</math> digits less than <math>2</math>.) | ||
+ | |||
+ | <math>\textbf{(A)} \, 500 \qquad\textbf{(B)} \, 625 \qquad\textbf{(C)} \, 1089 \qquad\textbf{(D)} \, 1199 \qquad\textbf{(E)} \, 1296 </math> | ||
==Solution== | ==Solution== |
Revision as of 10:40, 12 November 2022
Contents
[hide]Problem
How many strings of length formed from the digits , , , , are there such that for each , at least of the digits are less than ? (For example, satisfies this condition because it contains at least digit less than , at least digits less than , at least digits less than , and at least digits less than . The string does not satisfy the condition because it does not contain at least digits less than .)
Solution
Denote by the number of -digit strings formed by using numbers , where for each , at least of the digits are less than .
We have the following recursive equation: and the boundary condition for any .
By solving this recursive equation, for and , we get
For and , we get
For and , we get
For and , we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By OmegaLearn using Complementary Counting
~ pi_is_3.14
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.