Solve Enjoyable Enigmas with Mr.E's Challenge

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Enigman said:
Another one I can't believe I haven't asked yet...
Its a fruity one too!
(though I don't know if Gad's still following the thread...)
In a closed circular room there are 31 people standing such that everyone can see everyone else. Your job is to place a pomegranate such that all except one can see the fruit.

I'll have a go...

One wall is a mirror. 30 people face the mirror, 1 person faces them with his back to the mirror. Place the pomegranate behind this person.
 
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No mirrors and no smoke...
as its a silly one I will give it to you anyway.
:smile:
my answer was:
place it on the person's head...:-p
Now the more serious one:
Find the smallest palindromic number which is a perfect square and has an even number of digits.
EDIT: not writing a computer program would be appreciated...
 
Enigman said:
Find the smallest palindromic number which is a perfect square and has an even number of digits.
Does it have to be palindromic in base ten?

[STRIKE]If not, then I'd say nine (binary: 1001), which is a perfect square of three (32 = 9, or if you prefer in binary, 11 × 11 = 1001), and has an even number of bits (does the term "digit" imply base ten? Hmm).[/STRIKE] Edit: Nevermind, see next post.
 
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Enigman said:
Find the smallest palindromic number which is a perfect square and has an even number of digits.
My new guess is four (ternary 11). It is a perfect square of two, (ternary, 2 × 2 = 11), it is palindromic in base 3, and has an even number of digits in base 3.
 
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Just base ten collinsmark...
(For a smallest number it doesn't seem really small enough...
6 digits
)
 
collinsmark said:
Darn it.

Btw collinsmark I did solved it with C++... the clause was just so that I could see if someone's able to solve without programming...I couldn't.
:blushing:
 
Enigman said:
Btw collinsmark I did solved it with C++... the clause was just so that I could see if someone's able to solve without programming...I couldn't.
:blushing:

Nor could I in base 10. Well, not without incrementally checking numbers manually. Computers are better at stuff like that.

Code: 
using System;

// This is a C# program that finds the
// smallest perfect square that
// is palindromic in base 10,
// and has an even number of
// digits in base 10.
namespace Enigman_Palindrome_thing
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = 2; // This is the root of the perfect square.
            
            // Now let's loop through the natural numbers to find
            // the palindromic, even digited, perfect square.
            bool foundIt = false;
            while(!foundIt)
            {
                int nSquared = n * n; // this is the number to check.
                string nSquaredString = nSquared.ToString(); // string form.

                // Let the console know which number we're checking.
                Console.WriteLine("Currently checking " + nSquaredString);

                // check if it has even number of digits. 
                if((nSquaredString.Length)%2 == 0)
                {
                    // If we're here it's even. Check if it's a palindrome.
                    bool palindromeFlag = true;
                    for(int i = 0; i<nSquaredString.Length/2; i++)
                    {
                        // If symetric characters are not equal, 
                        // then it's not a palindrome.
                        if (!(nSquaredString[i] 
                            == nSquaredString[nSquaredString.Length - 1 - i]))
                            palindromeFlag = false;
                    }
                    if(palindromeFlag)
                    {
                        // Found it! let's report it to screen
                        // and finish up. 
                        foundIt = true;
                        Console.WriteLine(Environment.NewLine
                            + "Found it! The number is "
                            + nSquaredString + " which is "
                            + n.ToString() + " squared.");
                    }
                }
                // Increment n for next number to check.
                n++;
            }
            // Keep console alive until ready to leave.
            Console.ReadKey();

            // Goodbye!
        }
    }
}
 
Okay, you probably missed giving the final answer amidst copying your code so:
836^2=698698
 
There are eight oranges in a box. How can you divide them between eight people so that each person gets one orange, and one orange is still left in the box?
The oranges must not be peeled or cut.
 
Enigman said:
Okay, you probably missed giving the final answer amidst copying your code so:
836^2=698698

Or more correctly ...
836^2=698896

I see (from a one-line REXX program) that the next one has 12 digits, but I didn't have the patience to find the third one!
 
Enigman said:
There are eight oranges in a box. How can you divide them between eight people so that each person gets one orange, and one orange is still left in the box?
The oranges must not be peeled or cut.

Is it OK if one person gets a box as well, then?

(Edited to use spoiler tags - I thought I was just asking for clarification of the question!).
 
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Jonathan Scott said:
Is it OK if one person gets a box as well, then?

Done!
Oh, and spoiler it. :wink:
 
What is the maximum number of kings that can be placed on a chess board so that none of them threatens any other?
EDIT: A better way to put it would be maximum number of independent kings...
 
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Is that so that they all can't move as moving each King would place itself into check, or that they are positioned in a way that there is a concievable pattern of movements such that all kings could make movements which would ultimately lead to a draw?
 
Travis_King said:
Is that so that they all can't move as moving each King would place itself into check, or that they are positioned in a way that there is a concievable pattern of movements such that all kings could make movements which would ultimately lead to a draw?

The only condition is that the largest number of (independent ie. each of a different player or colour) kings should be on board without any king being in 'check' in the arrangement. So, the former follows from the conditions.
See- https://www.physicsforums.com/showpost.php?p=4575363&postcount=754 (same thing with knights)
This one too is a variation of the 8 queen puzzle. (Find an orientation such that there are 8 queens on board without being threatened ('checked') by any other)

A more mathematical restatement would be- in 8x8 square place maximum number of pieces so that there is always more than (or equal to) 1 unit^2 space between two pieces. or something like that...
 
Enigman said:
What is the maximum number of kings that can be placed on a chess board so that none of them threatens any other?
EDIT: A better way to put it would be maximum number of independent kings...
By trial and error only:
16?
 
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(I deleted the previous one posted here on grounds of being too silly)
If it takes four men eight days to dig four holes, how long does it
take one man to dig half a hole?
 
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Enigman said:
(I deleted the previous one posted here on grounds of being too silly)
If it takes four men eight days to dig four holes, how long does it
take one man to dig half a hole?

I find the concept of "half a hole" rather difficult.

Is this meant to be a trivial maths question where the holes are assumed to require a fixed amount of work and "half a hole" means half the amount of work?
 
The problem statement is all you get.
:biggrin:
(Okay, a leetle bit of advice: Do not stay in the box, stay and your fate's as uncertain as the cat's and I do love killing that particular variety of cat...I am a curious person)
 
I want to say that it's not possible to dig half a hole. Once you break dirt you've got yourself a whole hole!
 
Travis_King said:
I want to say that it's not possible to dig half a hole. Once you break dirt you've got yourself a whole hole!

Well said!
 
Here's a pretty easy one:

Two boys, Mikey and Jimmy are running a 100 meter foot-race. The first time they race Mikey beats Jimmy by 5 meters. To make things fair, the next time they race, Mikey stands 5 meters behind the starting line.

Assuming they each run just as fast as they did in the first race, does Jimmy win the second race?
 
AlephZero said:
64, if they are all black or all white :biggrin:
That's why I included 'independent' in the edit.
:smile:
Travis_King said:
Here's a pretty easy one:

Two boys, Mikey and Jimmy are running a 100 meter foot-race. The first time they race Mikey beats Jimmy by 5 meters. To make things fair, the next time they race, Mikey stands 5 meters behind the starting line.

Assuming they each run just as fast as they did in the first race, does Jimmy win the second race?

Let t be time taken by M
Vm-100/t velocity of M
Vj-95/t velocity of J
Now-
Tm=105/Vm
Tm=1.05t
Tj=100/Vj
Tj=1.053t
Tj>Tm
M wins
 
CompuChip - yup, you sure can.

That's it Enigman
some basic algebra for everyone!

How about this one:

A cashier tells you that the candybar you are trying to purchase costs $1.50. You hand him thirty five-cents. How much change do you get back?
 

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