Difference between revisions of "2022 AMC 12B Problems/Problem 23"
m (→See Also) |
Numerophile (talk | contribs) m (→Problem) |
||
Line 5: | Line 5: | ||
<cmath>S_n = \sum_{k=0}^{n-1} x_k 2^k</cmath> | <cmath>S_n = \sum_{k=0}^{n-1} x_k 2^k</cmath> | ||
− | Suppose <math>7S_n \equiv 1 \pmod{2^n}</math> for all <math>n \ | + | Suppose <math>7S_n \equiv 1 \pmod{2^n}</math> for all <math>n \geq 1</math>. What is the value of the sum |
<cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}</cmath> | <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}</cmath> | ||
Revision as of 22:31, 18 November 2022
- The following problem is from both the 2022 AMC 12B #23 and 2022 AMC 10B #25, so both problems redirect to this page.
Problem
Let be a sequence of numbers, where each
is either
or
. For each positive integer
, define
Suppose for all
. What is the value of the sum
Solution
First, notice that
Then since is the modular inverse of 7 in
, we can perform the Euclidean algorithm to find it for
.
Starting with ,
Now, take both sides
Using Fermat's Little Theorem,
Thus,
Therefore,
We may repeat this same calculation with to yield
Now, we notice that
is basically an integer expressed in binary form with
bits.
This gives rise to a simple inequality,
Since the maximum possible number that can be generated with
bits is
Looking at our calculations for
and
, we see that the only valid integers that satisfy that constraint are
~
Solution 2 (Base-2 Analysis)
We solve this problem with base 2.
To avoid any confusion, for a base-2 number, we index the th rightmost digit as digit
.
We have .
In the base-2 representation, is equivalent to
In the rest of the analysis, to lighten notation, we ease the base-2 subscription from all numbers. The equation above can be reformulated as:
\begin{table} \begin{tabular}{ccccccccc}
&& 0 &
& 0 & 0 & 0 & 0 & 0 \\ & & & & & & & & 1 \\
& &
&
&
&
&
&
&
\\ \hline %or \bottomrule if using the `booktabs` package &
![]()
![]()
&
&
&
&
& 0 & 0 & 0\\ \end{tabular}
\end{table}
Therefore, ,
, and for
,
.
Therefore,
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last problem | |
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 22 |
Followed by Problem 24 |
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.