2023 AIME I Problems/Problem 10
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
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 |
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