Difference between revisions of "2023 AIME I Problems/Problem 4"
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The induction fails starting at <math>n = 9</math> ! | The induction fails starting at <math>n = 9</math> ! | ||
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The actual answers <math>f(n)</math> for small <math>n</math> are: | The actual answers <math>f(n)</math> for small <math>n</math> are: | ||
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<math>0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12</math> | <math>0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12</math> | ||
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In general, <math>f(p) = f(p-1)+1</math> if p is prime, <math>n=4,6,8</math> are "lucky", and the pattern breaks down after <math>n=8</math> | In general, <math>f(p) = f(p-1)+1</math> if p is prime, <math>n=4,6,8</math> are "lucky", and the pattern breaks down after <math>n=8</math> | ||
Revision as of 17:28, 28 December 2024
Contents
[hide]Problem
The sum of all positive integers such that
is a perfect square can be written as
where
and
are positive integers. Find
Video Solution by MegaMath
https://www.youtube.com/watch?v=EqLTyGanr4s&t=136s
Solution 1
We first rewrite as a prime factorization, which is
For the fraction to be a square, it needs each prime to be an even power. This means must contain
. Also,
can contain any even power of
up to
, any odd power of
up to
, and any even power of
up to
. The sum of
is
Therefore, the answer is
.
~chem1kall
Solution 2
The prime factorization of is
To get
a perfect square, we must have
, where
,
,
.
Hence, the sum of all feasible is
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Educated Guess and Engineer's Induction (Fake solve))
Try smaller cases. There is clearly only one that makes
a square, and this is
. Here, the sum of the exponents in the prime factorization is just
. Furthermore, the only
that makes
a square is
, and the sum of the exponents is
here. Trying
and
, the sums of the exponents are
and
. Based on this, we (incorrectly!) conclude that, when we are given
, the desired sum is
. The problem gives us
, so the answer is
.
-InsetIowa9
However...
The induction fails starting at !
The actual answers
for small
are:
In general,
if p is prime,
are "lucky", and the pattern breaks down after
-"fake" warning by oinava
Solution 4
We have for some integer
. Writing
in terms of its prime factorization, we have
For a given prime
, let the exponent of
in the prime factorization of
be
. Then we have
Simplifying the left-hand side, we get
Thus, the exponent of each prime factor in
is at least
. Also, since
is prime and appears in the prime factorization of
, it follows that
must divide
.
Thus, is of the form
for some nonnegative integer
. There are
such values of
. The sum of all possible values of
is
The first sum can be computed using the formula for the sum of the first
squares:
Using the formula for the sum of a geometric series, we can simplify this as
The second sum can be computed using the formula for the sum of a geometric series:
Thus, the sum of all possible values of
is
so
.
- This answer is incorrect.
Video Solutions
I also somewhat try to explain how the formula for sum of the divisors works, not sure I succeeded. Was 3 AM lol. https://youtu.be/MUYC2fBF2U4
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.