1994 IMO Problems/Problem 3
Problem
For any positive integer , let be the number of elements in the set whose base 2 representation has precisely three s.
- (a) Prove that, for each positive integer , there exists at least one positive integer such that .
- (b) Determine all positive integers for which there exists exactly one with .
Solution
Solution 1
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.
Solution 2
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 and define the sets . So is the number of T-numbers in .
A positive even integer 2n is T iff n is T.(Fundamental theorem of T numbers)(FTT)
The function f is non-decreasing. It is sufficient to prove that . This follows by (FTT) if k+1 is T then 2k+2 is T too; so passing from to we can loose a T-number but we regain it with in so cannot be less than . We also have . This would be false if were not T and both and were T. But if were T then would be T too by the (FTT). Then we have that is or . But and so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values.
b) one-from-one values of f
The one-from-one values are such that and since f is monotone non-decreasing and has step 1.
By the condition since is T iff is T we have that must be T.
By the condition since is T iff is T we have that must be T.
Then and must have the form with since they are odd and since there is only 1 T-number less than 8 and they must have the same number of bits since their difference is 2 and both are T (the only two binary numbers which differs by 2 and have different number of bits are and or and which are evidently not T). Let and with . Then that is and . We conclude that is one-from-one for with . Since we have that with are the only one-from-one values of f.
See Also
1994 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |