Difference between revisions of "2022 AMC 10B Problems/Problem 25"
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-Rutvik Arora (youtube channel: https://www.youtube.com/channel/UCkgAgmNAQV8WGTOazGYEGwg) | -Rutvik Arora (youtube channel: https://www.youtube.com/channel/UCkgAgmNAQV8WGTOazGYEGwg) | ||
-whatdohumanitarianseat made a small edit for a typo | -whatdohumanitarianseat made a small edit for a typo | ||
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+ | ==Solution 5 == | ||
==Video Solution by MOP 2024== | ==Video Solution by MOP 2024== |
Revision as of 20:29, 27 May 2024
- The following problem is from both the 2022 AMC 10B #25 and 2022 AMC 12B #23, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Video Solution by MOP 2024
- 8 Video Solution by ThePuzzlr
- 9 Video Solution by Steven Chen
- 10 Video Solution by OmegaLearn Using Binary and Modular Arithmetic
- 11 Video Solution by The Power of Logic
- 12 Video Solution by Interstigation
- 13 See Also
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 1
In binary numbers, we have
It follows that
We obtain
by subtracting the equations:
We work from right to left:
For all
we conclude that
if and only if
if and only if
Finally, we get from which
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM
Solution 2
First, notice that
Then since
is the modular inverse of
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
.
~zoomanTV
Solution 3
As in Solution 2, we note that
We also know that
and
, this implies:
Dividing by
, we can isolate the previous sums:
The maximum value of
occurs when every
is equal to
. Even when this happens, the value of
is less than
. Therefore, we can construct the following inequalities:
From these two equations, we can deduce that both
and
are less than
.
Reducing and
we see that
and
The powers of repeat every
Therefore, and
Substituing this back into the above equations,
and
Since and
are integers less than
, the only values of
and
are
and
respectively.
The requested sum is
-Benedict T (countmath1)
Solution 4
Note that, as in Solution 2, we have This is because
and
Note that
Therefore,
Multiplying both sides by 7 gives us
We can write
and
for some a and b. Substituting, we get
Therefore, our answer can be written as
Another thing to notice is that a and b are integers between 0 and 6. This is because
which is
which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise,
which would mean
which cannot happen, so a is greater than 0. A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then and
so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is
-Rutvik Arora (youtube channel: https://www.youtube.com/channel/UCkgAgmNAQV8WGTOazGYEGwg) -whatdohumanitarianseat made a small edit for a typo
Solution 5
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution by ThePuzzlr
~ ThePuzzlr
Video Solution by Steven Chen
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Binary and Modular Arithmetic
~ pi_is_3.14
Video Solution by The Power of Logic
Video Solution by Interstigation
~Interstigation
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.