Difference between revisions of "2022 AMC 12A Problems/Problem 24"
(→Solution 2 (Cheese)) |
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==Solution 2 (Cheese)== | ==Solution 2 (Cheese)== | ||
Let the set of all valid sequences be <math>S</math>. | Let the set of all valid sequences be <math>S</math>. | ||
− | Notice that for any sequence <math>\{ | + | Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \{ | + | \{a_2,a_3,a_4,a_5,a_1\}\\ |
− | \{ | + | \{a_3,a_4,a_5,a_1,a_2\}\\ |
− | \{ | + | \{a_4,a_5,a_1,a_2,a_3\}\\ |
− | \{ | + | \{a_5,a_1,a_2,a_3,a_4\} |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | must also belong in <math>S</math>. The only exception is when all 5 elements are the same, i.e. <math>\{0,0,0,0,0\}</math>. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which by inspection of the answer choices yields <math>\boxed{(E) 1296}</math>. | + | must also belong in <math>S</math>. The only exception is when all 5 elements are the same, i.e. <math>\{0,0,0,0,0\}</math>. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which by inspection of the answer choices yields <math>\boxed{\textbf{(E) 1296}}</math>. |
~Tau | ~Tau |
Revision as of 13:49, 12 November 2022
Contents
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)
Solution 2 (Cheese)
Let the set of all valid sequences be .
Notice that for any sequence
in
, the sequences
must also belong in
. The only exception is when all 5 elements are the same, i.e.
. Then
, which by inspection of the answer choices yields
.
~Tau
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.