# Logic Problems for Pre-RMO and RMO practice

Problem 1:

Peter’s mom said: “All champions are good at math.” Peter says: “I am good at math. Therefore, I am champion!” Is his implication right or wrong?

Problem 2:

There are four cards on the table with the symbols A, B, 4, and 5 written on their visible sides. What is the minimum number of cards we must turn over to find out whether the following statement is true: “If an even number is written on one side of a card, then a vowel is written on the other side?”

Problem 3:

A sum of 15 cents was paid by two coins, and one of these coins was not a nickel. Find the values of the coins.

Problem 4:

Assume that the following statements are true:

(a) among people having TV sets there are some who are not mathematicians.

(b) non-mathematicians who swim in swimming pools every day do not have TV sets.

Can we claim that not all people having TV sets swim every day?

Problem 5:

During a trial in Wonderland the March Hare claimed that the cookies were stolen by Mad Hatter. Then, the Mad Hatter and the Dormouse gave testimonies which, for some reason, were not recorded. Later on in the trial, it was found out that the cookies were stolen by only one of these three defendants, and, moreover, only the guilty one gave true testimony. Who stole the cookies?

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More later,

Nalin Pithwa

# A gem of an inequality from USSR !!!

Problem:

The positive numbers a, b, c, A, B, C satisfy

$a+A=b+B=c+C=k$.

Prove that $aB+bC+cA \leq k^{2}$.

Solution:

Proving that $aB+bC+cA \leq k^{2}$ is equivalent to showing that

$aB+bC+cA = a(k-b)+b(k-c)+c(k-a) \leq k^{2}$

$\Longleftrightarrow k(a+b+c) \leq k^{2}+ab+bc+ca$

But this is equivalent to

$k^{2}-ka+bc \geq (b+c)(k-a) \Longleftrightarrow bc \geq (k-a)(b+c-k)$

If $b+c-k \leq 0$, then the above is obviously true. If not, then we have

$(k-b)(k-c) \geq 0 \Rightarrow bc \geq k(b+c-k) \geq (k-a)(b+c-k)$

and thus, our inequality holds.

More later,

Nalin Pithwa

# A cute problem on permutations and combinations

Here, is a problem on counting. Read it and crack it before you compare your solution with the one I present !!!

In a tennis tournament, there are $2n$ participants. In the  first round of  the tournament, each participant plays just once, so there are n games, each occupying a pair of  players. Show that the pairing for the first round can be arranged in exactly

$1 x 3 x 5 x 7 x 9 \ldots(2n-1)$

different ways.

Solution.

Call the required number of pairings of $2n$ players $P_{n}$. If you are a participant, you can be matched with any one of  the other $(2n-1)$ players. Once your antagonist is chosen, there remain

$(2n-2)=2(n-1)$

players who  can be paired in $P_{n-1}$ ways. Hence,

$P_{n}=(2n-1)P_{n-1}$.