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# Queen shock

• #41

This chatbox is not fit for mathematical reasoning but the solution I find would be

4096 x 3 x 25

4096 x 3 x 49

4096 x 7 x 25

4096 x 9 x 5

4096 x 9 x 7

4096 x 9 x 11

4096 x 9 x 13

4096 x 9 x 17

4096 x 9 x 19

4096 x 9 x 23

• #42

The reasoning goes as follows

Every number can be written as p1^a1 x ..... pk^ak the pi being different primes, a1,..,ak at least 1.

The numberof divisors for such a number is (a1+1) x.....x(ak+1) - 1.

In order this to be 77 the product should be 78 = 2x3x13.

This limits the choices for k (being at most 3) and a1,a2 and a3.

I leave it as an excersise to see why k=1 or k=2 are impossible... the numbers would become tobig even if one picks the smallest two primes.

So k=3 a1=1, a2=2 and a3 = 12.

ITis easy to see that the listed possibilities are the only ones satisfying all conditions....

• #43

Strangely, I got a different answer. It went like this:

First of all, if a natural number is to have an odd number of divisors, it must be a perfect square.

Why? For a non-square natural number, its divisors can be written in the form of a number of pairs a * b where a < sqrt n < b.

Secondly, any natural number can be written in the form (p1^a1)(p2^a2)(p3^a3).... in which case the number of divisors is equal to (a1+1)(a2+1)(a3+1)... (I assume the difference between our answers is because I counted both 1 and n as divisors of n whereas you chose not to count one of those and got different results.) 77 = 11*7, so k=2, a1=10 and a2=6.

That leaves us with two possible answers:

3^10 x 2^6 = 3,779,136 > 1,000,000

2^10 x 3^6 = 746,496

• #44

Oh yes i was thinking to complicated again, yes 1 should be counted as a divisor...... and i didn t do that....

• #45

Still you have to consider the case k=1... but the numbers would be to big.....

• #46

Whatever. It's enough that we can debate something respectfully on the Internet That's yet another great thing about math. Nobody cares about it enough to start a flame war

• #47

All of them, isn't it? I mean, it's kind of like the "which months have 28 days riddle."

• #48

Yes, but i forgot to count 1 as a divisor ......

• #49

So actually I solved two problems

1) Determine the numbers less then 1,000,000 having exactly 77 divisors bigger then 1

2) Determine the numbers less then 1,000,000 having exactly 77 real divisors (i.e. divisors not being the number itself)

• #50

Well, yes. I would ask if you have any other fun ones, but I don't want to derail the thread, so we should get back to the OP's game.

• #51

Oh yes there is a quite nice one, the solution can be found on the internet but it is more fun to find it yourself.

There are hundred prisoners that are going to be executed unless they survive the following game.

In a room the numbers 1-100 are stored in 100 drawers each containing 1 number.

The prisoners are numbered 1-100. They consecutively have to enter the room and may open 50 drawers.

If every prisoner manages to open a drawer containing his own number they will survive, if one prisoner does not succeed finding his own number they all are executed. The drawers are closed again after a prisoner leaves the room. They may discuss a strategy before the game is played but during the game they cannot communicate. Is there a strategy which gives them a reasonable chance of survival?

• #52

Just to clarify: 50 drawers each, or 1 drawer each for 50 prisoners?

• #53

50 drawers each I assume

• #54

Okay. That's awfully tough, as then the chance of survival without a strategy is 1/2^100.

• #55

There are a total of 100 drawers each contains one number in the range 1-100, all numbers are present.Each prisoner may open at most 50 drawers in search of his own number. The drawers re closed again when a prisoner leaves the room. During the game prisoners cannot communicate. There are a total of hundred prisoner. Each one has to find his own number.

• #56

Nope , there is a strategy giving approx chance 0.31 that they survive.

• #57

Now find the strategy. Tip... some knowledge of combinatorics and group theory is equired.

• #58

Sadly, I've never done anything with combinatorics...

• #59

The strategy is as follows, The prisoners number the drawers from left to right and top to bottem , so every drawer now is assigned a number at the outside and contains a number in the insiden. The prisoners assign themselves numbers 1 up to 100.

When prisoner number k enters the room he starts opening drawer with outside number k. He inspects the content and either find his own number or a number say i and then opens the drawer with outside number i.

He keeps opening drawers based on the inspected content tlll he finds his own number or opened 50 drawers.

Now you may figure out yourself why this gives a relatively high probability of survival. It has to do with permutation containing not to long cycles.....

• #60
Alika05 wrote:

I'll give a trophy to the first person who can find a game someone showcased in which they lost. Happy Hunting.

See comment #37. Now where's my trophy?

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