# Difference between revisions of "1994 IMO Problems/Problem 3"

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But <math>f(4)=1</math> and <math>f(2^n)= \tbinom n2</math> so f takes all of the positive integer values because starting from 1 with step of 1 reaches arbitrarily large integer values. | But <math>f(4)=1</math> and <math>f(2^n)= \tbinom n2</math> 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 <math>f(k)</math> are such that <math>f(k)=f(k-1)+1</math> and <math>f(k+1)=f(k)+1</math> since f is monotone non-decreasing and has step 1. | The one-from-one values <math>f(k)</math> are such that <math>f(k)=f(k-1)+1</math> and <math>f(k+1)=f(k)+1</math> since f is monotone non-decreasing and has step 1. |

## Latest revision as of 14:36, 1 July 2012

## Contents

## Problem

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

### 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 diferrence is 2 and both are T (the only two binary numbers wich 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.