Difference between revisions of "1994 IMO Problems/Problem 3"
(→a) Surjectivity of f) |
(→b) one-from-one values) |
||
Line 41: | Line 41: | ||
By the fact that <math>f(2^n +2^r + 2^s)=f(2^n) + r +1 </math> where <math>r=1,2,...,n-1</math> and <math>r > s \ge 0</math> | By the fact that <math>f(2^n +2^r + 2^s)=f(2^n) + r +1 </math> where <math>r=1,2,...,n-1</math> and <math>r > s \ge 0</math> | ||
− | it follows that the values <math>f(2^n) + r +1</math> where <math>r=2,...,n-1</math> come from different k because there are at least two different choices for <math>s</math>. These values are <math>f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})</math>. | + | it follows that each of the values <math>f(2^n) + r +1</math> where <math>r=2,...,n-1</math> come from at least two different k because there are at least two different choices for <math>s</math>. These values are <math>f(2^n)+3,f(2^n)+4,...,f(2^n)+n=f(2^{n+1})</math>. |
Thus the only possible one-from-one values are <math>f(2^n)+1</math> that come from <math>k=2^n + 2</math> and <math>f(2^n)+2</math> that come from <math>k=2^n + 3</math>. | Thus the only possible one-from-one values are <math>f(2^n)+1</math> that come from <math>k=2^n + 2</math> and <math>f(2^n)+2</math> that come from <math>k=2^n + 3</math>. |
Revision as of 20:58, 31 October 2009
3. For any positive integer k, let f(k) be the number of elements in the set {k + 1; k + 2;....; 2k} whose base 2 representation has precisely three 1s.
- (a) Prove that, for each positive integer m, there exists at least one positive integer k such that f(k) = m.
- (b) Determine all positive integers m for which there exists exactly one k with f(k) = m.
Solution
a) Surjectivity of f
For space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.
It's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number)
whence
Now consider with
and the corrisponding set
whose subset
contains
T-numbers since
contains
T-numbers by definition but
has none. So
but the last set has the only T-number
. We conclude that:
Now consider with
and
.
We explicitly calculate f(k) for such numbers.
So we have to calculate how many T-numbers are in the set
.
The T-numbers less than in
are
whence
.
The T-numbers greater than in
are
whence
.
Therefore contains
T-numbers and we have:
where
ans
.
Summarizing
where
and
.
That's to say that f takes all the values from to
for every
and
then takes any positive integer value since
and
becomes arbitrarily large.
b) one-from-one values
By the fact that where
and
it follows that each of the values
where
come from at least two different k because there are at least two different choices for
. These values are
.
Thus the only possible one-from-one values are that come from
and
that come from
.
If we have
because k+1 is not T and
and
are not T so f maps
and
to
.
If then
.
The function f is non-decreasing.
It is sufficient to prove that
but this follows by the fact that if k+1 is T then 2k+2 is T too.
By the monotonicity of f
with
are the only one-from-one values of f.