Difference between revisions of "1987 IMO Problems/Problem 4"
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Suppose <math>f(m) = n</math> with <math>m \equiv n \mod k</math>. Then by an easy induction on <math>r</math> we find <math>f(m + kr) = n + kr</math>, <math>f(n + kr) = m + k(r+1)</math>. We show this leads to a contradiction. Suppose <math>m < n</math>, so <math>n = m + ks</math> for some <math>s > 0</math>. Then <math>f(n) = f(m + ks) = n + ks</math>. But <math>f(n) = m + k</math>, so <math>m = n + k(s - 1) \ge n</math>. Contradiction. So we must have <math>m \ge n</math>, so <math>m = n + ks</math> for some <math>s \ge 0</math>. But now <math>f(m + k) = f(n + k(s+1)) = m + k(s + 2)</math>. But <math>f(m + k) = n + k</math>, so <math>n = m + k(s + 1) > n</math>. Contradiction. | Suppose <math>f(m) = n</math> with <math>m \equiv n \mod k</math>. Then by an easy induction on <math>r</math> we find <math>f(m + kr) = n + kr</math>, <math>f(n + kr) = m + k(r+1)</math>. We show this leads to a contradiction. Suppose <math>m < n</math>, so <math>n = m + ks</math> for some <math>s > 0</math>. Then <math>f(n) = f(m + ks) = n + ks</math>. But <math>f(n) = m + k</math>, so <math>m = n + k(s - 1) \ge n</math>. Contradiction. So we must have <math>m \ge n</math>, so <math>m = n + ks</math> for some <math>s \ge 0</math>. But now <math>f(m + k) = f(n + k(s+1)) = m + k(s + 2)</math>. But <math>f(m + k) = n + k</math>, so <math>n = m + k(s + 1) > n</math>. Contradiction. | ||
− | So if <math>f(m) = n</math>, then <math>m</math> and <math>n</math> have different residues <math>\ | + | So if <math>f(m) = n</math>, then <math>m</math> and <math>n</math> have different residues <math>\pmod k</math>. Suppose they have <math>r_1</math> and <math>r_2</math> respectively. Then the same induction shows that all sufficiently large <math>s \equiv r_1 \pmod k</math> have <math>f(s) \equiv r_2 \pmod k</math>, and that all sufficiently large <math>s \equiv r_2 \pmod k</math> have <math>f(s) \equiv r_1 \pmod k</math>. Hence if <math>m</math> has a different residue <math>r \mod k</math>, then <math>f(m)</math> cannot have residue <math>r_1</math> or <math>r_2</math>. For if <math>f(m)</math> had residue <math>r_1</math>, then the same argument would show that all sufficiently large numbers with residue <math>r_1</math> had <math>f(m) \equiv r \pmod k</math>. Thus the residues form pairs, so that if a number is congruent to a particular residue, then <math>f</math> of the number is congruent to the pair of the residue. But this is impossible for <math>k</math> odd. |
==Other Solution== | ==Other Solution== | ||
− | Solution by | + | Solution by Sawa Pavlov: |
− | Let <math>N</math> be the set of non-negative integers. Put <math>A = N - f(N)</math> (the set of all n such that we cannot find m with f(m) = n). Put <math>B = f(A)</math>. | + | Let <math>N</math> be the set of non-negative integers. Put <math>A = N - f(N)</math> (the set of all <math>n</math> such that we cannot find m with f(m) = n). Put <math>B = f(A)</math>. |
Note that <math>f</math> is injective because if <math>f(n) = f(m)</math>, then <math>f(f(n)) = f(f(m))</math> so <math>m = n</math>. We claim that <math>B = f(N) - f(f(N))</math>. Obviously <math>B</math> is a subset of <math>f(N)</math> and if <math>k</math> belongs to <math>B</math>, then it does not belong to <math>f(f(N))</math> since <math>f</math> is injective. Similarly, a member of <math>f(f(N))</math> cannot belong to <math>B</math>. | Note that <math>f</math> is injective because if <math>f(n) = f(m)</math>, then <math>f(f(n)) = f(f(m))</math> so <math>m = n</math>. We claim that <math>B = f(N) - f(f(N))</math>. Obviously <math>B</math> is a subset of <math>f(N)</math> and if <math>k</math> belongs to <math>B</math>, then it does not belong to <math>f(f(N))</math> since <math>f</math> is injective. Similarly, a member of <math>f(f(N))</math> cannot belong to <math>B</math>. |
Revision as of 02:12, 15 March 2008
Problem
Prove that there is no function from the set of non-negative integers into itself such that
for every
.
Solution
We prove that if for all
, where
is a fixed positive integer, then
must be even. If
, then we may take
.
Suppose with
. Then by an easy induction on
we find
,
. We show this leads to a contradiction. Suppose
, so
for some
. Then
. But
, so
. Contradiction. So we must have
, so
for some
. But now
. But
, so
. Contradiction.
So if , then
and
have different residues
. Suppose they have
and
respectively. Then the same induction shows that all sufficiently large
have
, and that all sufficiently large
have
. Hence if
has a different residue
, then
cannot have residue
or
. For if
had residue
, then the same argument would show that all sufficiently large numbers with residue
had
. Thus the residues form pairs, so that if a number is congruent to a particular residue, then
of the number is congruent to the pair of the residue. But this is impossible for
odd.
Other Solution
Solution by Sawa Pavlov:
Let be the set of non-negative integers. Put
(the set of all
such that we cannot find m with f(m) = n). Put
.
Note that is injective because if
, then
so
. We claim that
. Obviously
is a subset of
and if
belongs to
, then it does not belong to
since
is injective. Similarly, a member of
cannot belong to
.
Clearly and
are disjoint. They have union
which is
. But since
is injective they have the same number of elements, which is impossible since
has an odd number of elements.
1987 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |