Difference between revisions of "2013 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | {{ | + | We prove the claim by induction on <math>k</math>. |
+ | |||
+ | '''Base case:''' If <math>k = 1</math> then <math>1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}</math>, so the claim is true for all positive integers <math>n</math>. | ||
+ | |||
+ | '''Inductive hypothesis:''' Suppose that for some <math>m \in \mathbb{Z}^{+}</math> the claim is true for <math>k = m</math>, for all <math>n \in \mathbb{Z}^{+}</math>. | ||
+ | |||
+ | '''Inductive step:''' Let <math>n</math> be arbitrary and fixed. Case on the parity of <math>n</math>: | ||
+ | |||
+ | [Case 1: <math>n</math> is even] <math>1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m} - 1}{\frac{n}{2}} \right) \cdot \left( 1 + \frac{1}{n + 2^{m+1} - 2} \right)</math> | ||
+ | |||
+ | [Case 2: <math>n</math> is odd] <math>1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m}-1}{\frac{n+1}{2}} \right) \cdot \left( 1 + \frac{1}{n} \right)</math> | ||
+ | |||
+ | In either case, <math>1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^m - 1}{c} \right) \cdot \left( 1 + \frac{1}{a_{m+1}} \right)</math> for some <math>c, a_{m+1} \in \mathbb{Z}^+</math>. | ||
+ | |||
+ | By the induction hypothesis we can choose <math>a_1, ..., a_m</math> such that <math>\left( 1 + \frac{2^m - 1}{c} \right) = \prod_{i=1}^{m} (1 + \frac{1}{a_i})</math>. | ||
+ | |||
+ | Therefore, since <math>n</math> was arbitrary, the claim is true for <math>k = m+1</math>, for all <math>n</math>. Our induction is complete and the claim is true for all positive integers <math>n</math>, <math>k</math>. | ||
+ | |||
+ | ==Alternative Solution== | ||
+ | We will prove the claim by constructing telescoping product out of positive integers: | ||
+ | <cmath>\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3}\cdot \cdot \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(n+2^{k}-1\right)}{n}</cmath> | ||
+ | where <math>a_1=n</math> and where each fraction <math>\frac{a_{i+1}}{a_i}=\frac{a_{i}+\Delta_i}{a_i}</math> can also be written as <math>\frac{m_i+1}{m_i}</math> for some positive integer <math>m_i</math>. All <math>\Delta_i</math> will be different with <math>\Delta_i=2^j</math>, <math>0\le j \le (k-1)</math>. <math>\sum_{i=1}^{k}{\Delta_i}=\sum_{i=1}^{k}{2^{i-1}}=2^{k}-1</math>. | ||
+ | |||
+ | '''Ascend:''' make telescoping product of fractions with a sequence of <math>\Delta_i</math> that increases in magnitude up to the maximum <math>2^{k-1}</math>. If the maximum <math>\Delta=2^{k-1}</math> is reached go to the descend phase. | ||
+ | Pull out factors of <math>2</math> (up to the maximum <math>2^{k-1}</math>). | ||
+ | <math>n=2^j(2z+1)</math>, <math>z+1=2^q(2r+1)</math> etc where <math>j\ge 0</math>, <math>q\ge 0</math> ... | ||
+ | Construct telescoping as <cmath>\frac{2^j(2z+2)}{2^j(2z+1)}\cdot \frac{2^{j+q+1}(2r+2)}{2^{j+q+1}(2r+1)} ...</cmath>. The corresponding differences <math>\Delta</math> are <math>2^j, 2^{j+q+1}</math> ... Every <math>\Delta_i</math> is bigger then the previous <math>\Delta_{i-1}</math> by at least factor <math>2</math>. Therefore, the biggest needed <math>\Delta=2^{k-1}</math> can be created telescoping at most <math>k</math> fractions. After we constructed the fraction <math>\frac{2^{k-1}(h+1)}{2^{k-1}h}</math> with the biggest needed <math>\Delta=2^{k-1}</math> we can construct any remaining needed <math>\Delta_i</math> as described below. | ||
+ | |||
+ | '''Descend:''' | ||
+ | If we need <math>\Delta=2^{d}</math> where <math>d<k-1</math> we can use as the next telescoping fraction <math>\frac{2^{d}(2^{k-1-d}(h+1)+1)}{2^{d}(2^{k-1-d}(h+1))}</math>. We can construct all the remaining nedded <math>\Delta_i</math> in the decreasing order of their magnitude by repeating the same step. | ||
+ | |||
+ | --[[User:alexander_skabelin|alexander_skabelin]] 9:24, 11 July 2023 (EST) | ||
+ | |||
+ | {{alternate solutions}} | ||
==See Also== | ==See Also== | ||
*[[2013 IMO]] | *[[2013 IMO]] | ||
+ | {{IMO box|year=2013|before=First Problem|num-a=2}} |
Latest revision as of 01:30, 19 November 2023
Problem
Prove that for any pair of positive integers and
, there exist
positive integers
(not necessarily different) such that
.
Solution
We prove the claim by induction on .
Base case: If then
, so the claim is true for all positive integers
.
Inductive hypothesis: Suppose that for some the claim is true for
, for all
.
Inductive step: Let be arbitrary and fixed. Case on the parity of
:
[Case 1: is even]
[Case 2: is odd]
In either case, for some
.
By the induction hypothesis we can choose such that
.
Therefore, since was arbitrary, the claim is true for
, for all
. Our induction is complete and the claim is true for all positive integers
,
.
Alternative Solution
We will prove the claim by constructing telescoping product out of positive integers:
where
and where each fraction
can also be written as
for some positive integer
. All
will be different with
,
.
.
Ascend: make telescoping product of fractions with a sequence of that increases in magnitude up to the maximum
. If the maximum
is reached go to the descend phase.
Pull out factors of
(up to the maximum
).
,
etc where
,
...
Construct telescoping as
. The corresponding differences
are
... Every
is bigger then the previous
by at least factor
. Therefore, the biggest needed
can be created telescoping at most
fractions. After we constructed the fraction
with the biggest needed
we can construct any remaining needed
as described below.
Descend:
If we need where
we can use as the next telescoping fraction
. We can construct all the remaining nedded
in the decreasing order of their magnitude by repeating the same step.
--alexander_skabelin 9:24, 11 July 2023 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2013 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |