Difference between revisions of "2021 AIME II Problems/Problem 15"
MRENTHUSIASM (talk | contribs) (→Solution 2 (More Variables)) |
MRENTHUSIASM (talk | contribs) m (→Solution 2 (More Variables)) |
||
Line 29: | Line 29: | ||
f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{15mm}(2) \\ | f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{15mm}(2) \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting with <math>g\left((k+1)^2\right)=k+1,</math> we | + | If <math>n</math> and <math>(k+1)^2</math> have the same parity, then starting with <math>g\left((k+1)^2\right)=k+1,</math> we get <math>g(n)=f(n)</math> by a similar process. This contradicts the precondition <math>\frac{f(n)}{g(n)} = \frac{4}{7}.</math> Therefore, <math>n</math> and <math>(k+1)^2</math> must have different parities, from which <math>n</math> and <math>(k+2)^2</math> must have the same parity. |
Along with the earlier restriction, note that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{15mm}(3)</cmath> for some positive integer <math>q.</math> By observations, we get | Along with the earlier restriction, note that <math>k^2<n<(k+2)^2,</math> or <cmath>n=(k+2)^2-2q\hspace{15mm}(3)</cmath> for some positive integer <math>q.</math> By observations, we get |
Revision as of 04:27, 12 May 2021
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 (More Variables)
We restrict in which
for some positive integer
or
for some nonnegative integer
By observations, we get
If
and
have the same parity, then starting with
we get
by a similar process. This contradicts the precondition
Therefore,
and
must have different parities, from which
and
must have the same parity.
Along with the earlier restriction, note that or
for some positive integer
By observations, we get
SOLUTION IN PROGRESS. NO EDIT PLEASE--A MILLION THANKS.
Remark
We can verify that
~MRENTHUSIASM
Video Solution
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.