Difference between revisions of "2023 AIME I Problems/Problem 10"
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<math>U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}</math> | <math>U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}</math> | ||
− | Where <math>n=2023</math>. Note that since <math>5 \cdot 404=2020</math>, this is an approximation for <math>U</math> because the equation disregards the remainder (mod <math>5</math>) when <math>n= | + | Where <math>n=2023</math>. Note that since <math>5 \cdot 404=2020</math>, this is an approximation for <math>U</math> because the equation disregards the remainder (mod <math>5</math>) when <math>n=2021, 2022, and 2023</math> so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of <math>U</math> (*). However, we will find that this is a negligible error when it comes to the inequality <math>-1000<U<1000</math>, so we can proceed with this approximation to solve for <math>a</math>. |
Factoring our approximation gives <math>U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}</math> | Factoring our approximation gives <math>U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}</math> |
Revision as of 16:32, 8 February 2023
Contents
Problem 10
There exists a unique positive integer for which the sum
is an integer strictly between
and
. For that unique
, find
.
(Note that denotes the greatest integer that is less than or equal to
.)
Solution (Bounds and Decimal Part Analysis)
Define .
First, we bound .
We establish an upper bound of . We have
We establish a lower bound of . We have
We notice that if , then
.
Thus,
Because and
, we must have either
or
.
For , we get a unique
.
For
, there is no feasible
.
Therefore, . Thus
.
Next, we compute .
Let , where
.
We have
Therefore,
Therefor, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Punxsutawney Phil
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4
We define . Since for any real number
,
, we have
. Now, since
, we have
.
Now, we can solve for in terms of
. We have:
So, we have
, and
, so we have
, or
. Now,
is much bigger than
or
, and since
is an integer, to satsify the inequalities, we must have
, or
, and
.
Now, we can find . We have:
Now, if
, then
, and if
, then
, and so on. Testing with
, we get
respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for
, we get:
Since
, this gives
, and we have
.
~ genius_007
Solution 5 (Quick)
We can view the floor function in this problem as simply subtracting the remainder of (mod
) from the numerator of
. For example,
.
Note that the congruence of (mod
) loops every time
increases by 5. Also, note that the congruence of
(mod
) determines the set of congruences of
for each congruence of
(mod
).
For example, if (mod
), the set of remainders is
for
(mod
). Let the sum of these elements be
. Note that for each “loop” of the numerator (mod
), each element of the set will be subtracted exactly once, meaning
is subtracted once for each loop. The value of the numerator will loop
times (mod
) throughout the sum, as
. Then
Where . Note that since
, this is an approximation for
because the equation disregards the remainder (mod
) when
so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of
(*). However, we will find that this is a negligible error when it comes to the inequality
, so we can proceed with this approximation to solve for
.
Factoring our approximation gives
We set to make
, accordingly minimizing
, yielding
If increases or decreases by
, then
changes by
which clearly breaks the inequality on
. Therefore
(mod
) giving the set of remainders
, so
and our approximation yields
. However, we must subtract 2, 1, and 2 (*) giving us
, giving an answer of
~Spencer Danese
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.