2021 AIME II Problems/Problem 15
Contents
Problem
Let and
be functions satisfying
and
for positive integers
. Find the least positive integer
such that
.
Solution 1
Consider what happens when we try to calculate where n is not a square. If
for (positive) integer k, recursively calculating the value of the function gives us
. Note that this formula also returns the correct value when
, but not when
. Thus
for
.
If ,
returns the same value as
. This is because the recursion once again stops at
. We seek a case in which
, so obviously this is not what we want. We want
to have a different parity, or
have the same parity. When this is the case,
instead returns
.
Write , which simplifies to
. Notice that we want the
expression to be divisible by 3; as a result,
. We also want n to be strictly greater than
, so
. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is
, giving
.
Indeed - if we check our answer, it works. Therefore, the answer is .
-Ross Gao
Solution 2 (Four Variables)
We consider and
separately:
We restrict
in which
for some positive integer
or
for some integer
such that
By recursion, we get
If
and
have the same parity, then starting from
we get
by a similar process. This contradicts the precondition
Therefore,
and
must have different parities, from which
and
must have the same parity.
It follows that
or
for some integer
such that
By recursion, we get
- Answer
By
and
we have
From
and
equating the expressions for
gives
Solving for
produces
We substitute
into
then simplify, cross-multiply, and rearrange:
Since
we know that
must be divisible by
and
must be divisible by
Recall that the restrictions on
and
are
and
respectively. Substituting
into either inequality gives
Combining all these results produces
To minimize
in either
or
we minimize
so we minimize
and
in
From
and
we construct the following table:
Finally, we have
Substituting this result into either
or
generates
- Remark
We can verify that
~MRENTHUSIASM
Solution 3
Since isn't a perfect square, let
with
. If
is odd, then
. If
is even, then
from which
Since
is even,
is even. Since
, the smallest
is
which produces the smallest
.
.
~Afo
Solution 4 (Quadratics With Two Variables)
To begin, note that if is a perfect square,
, so
, so we must look at values of
that are not perfect squares (what a surprise). First, let the distance between
and the first perfect square greater than or equal to it be
, making the values of
and
integers. Using this notation, we see than
, giving us a formula for the numerator of our ratio. However, since the function of
does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of
in
unless
is an even number. However, this is impossible, since if
was an even number,
, giving a ratio of one. Thus,
must be an odd number.
Thus, since must be an odd number, regardless of whether
is even or odd, to get an integral value in
, we must get to the next perfect square after
. To make matters easier, let
. Thus, in
, we want to achieve
.
Expanding and substituting in the fact that
yields:
Thus, we must add the quantity to
to achieve a integral value in the function
. Thus.
However, note that the quantity within the square root is just , and so:
Thus,
Since we want this quantity to equal , we can set the above equation equal to this number and collect all the variables to one side to achieve
Substituting back in that , and then separating variables and squaring yields that
Now, if we treat as a constant, we can use the quadratic formula in respect to
to get an equation for
in terms of
(without all the squares). Doing so yields
Now, since and
are integers, we want the quantity within the square root to be a perfect square. Note that
. Thus, assume that the quantity within the root is equal to the perfect square,
. Thus, after using a difference of squares, we have
Since we want
to be an integer, we know that the
should be divisible by five, so, let's assume that we should have
divisible by five. If so, the quantity
must be divisible by five, meaning that
leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).
Thus, we see that to achieve integers and
that could potentially satisfy the problem statement, we must try the values of
congruent to one modulo five. However, if we recall a statement made earlier in the problem, we see that we can skip all even values of
produced by this modulo argument.
Also, note that won't work, as they are too small, and will give an erroneous value for
. After trying
, we see that
will give a value of
, which yields
, which, if plugged in to for our equations of
and
, will yield the desired ratio, and we're done.
Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with "needs proof". If you can prove it, please add a lemma to the solution doing so :)
-mathislife52
Video Solution
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
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