Interesting problems

[brightsky]

From past papers of various maths competitions:

1. Prove that the number (with at least one zero) is not a perfect square.

2. Prove that among 18 consecutive three digit numbers there must be at least one which is divisible by the sum of its digits.

3. Do there exist functions and , such that is an even function while is an odd function (different from 0)?

4. Quadrilateral is a cyclic, . Points and are chosen on sides and respectfully so that . Prove that .

5. Can it happen that the least common multiple of is times the least common multiple of for some positive integers and ?

6. Each term of an infinite sequence of natural numbers is obtained from the previous term by adding to it one of its nonzero digits. Prove that this sequence contains an even number.

7. Let be any three numbers from an open interval . Prove the inequality:
is less than or equal to

[kamil9876]

5 looks like my sort of problem

let A={1,2,3...m}, B={1,2,3...n}

therefore in order for the condition to be satisfied we require the following:

1.) if is the largest power of 2 in A, then must be the largest power of 2 in B.

2.) if is the largest powe of in A, then must be the largest power of 251 in B.

3.) No power of a prime other than 2 (e.g: like ) may be in B/A (B without A) since then the ratio of the LCM's would then be a multiple of 3, and 2008 is not a multiple of 3.

We will show that these conditions cannot be satisfied simultaenously by showing that if 1 is satisfied, then 3 is not. To prove this it suffices to prove that there is a power of 3 between and and thus a power of 3 in A/B, which shows condition 3 may not hold.

This statement can be proven by the following result: for every positive integer , there exists a such that:



We will prove this by induction. It's obviously true for M=1.

Now how suppose that . if then the inductive step is completed, if not then:



This implies two things:

a.) and
b.)

Combining a and b we get: which completes the proof.

Which shows that the answer to the question is "NO, we cannot"

[brightsky]

4. Extend from point to point such that .

Draw line .

quadrilateral is cyclic, hence (opposite angles of a cyclic quadrilateral add to 180 degrees).

(angles on a straight line add to 180 degrees).



In triangle and triangle ,

(given), and

(SAS)

Hence and (corresponding angles and sides of congruent triangle are equal)

So

In triangle and triangle ,

and (common).

(SAS)

Hence (correponding sides of congruent triangle are equal).

So