A doctor adviced his patient to walk 10 Km every day.

After one month he phoned the Doctor :Doc. I am now near the city border! shall I go further?

As discussed, this is simply a case of a systems of equations: M + W + C = 40 and 4M + 2W + C/4 = 40.

Since there are only two equations and three variables, there are potentially infinite solutions but constraints limit the possibilities.

As noted, each variable must be an integer (no half people) - and nonnegative.

Also, since the final amount paid is an integer, the number of children must be a multiple of four. (10 possibilities)

Furthermore, since both the men's and women's costs are even, the number of children must be a multiple of eight. (5 possibilities)

Finally, the remaining amount of money after taking out the costs for children must be no more than $4 per remaining person (all men) or less than $2 per remaining person (all women).

That leaves only the two remaining possibilities: 8 men, 0 women, and 32 children or 1 man, 15 women, and 24 children.

Interestingly, one could argue that there are 2, 1 or 0 solutions based on different interprtations of the problem:

if "men, women, and children" implies there must be at least one of each, only 1 solution remains;

if it implies there must be at least two of each, no solutions remain.

BTW, without the restriction on the total number of people, there are 110 valid solutions if any variable can be either 0 or 1 and

only 56 solutions if there must be at least two men, two women, and two children.

the solution....1 man...15 women....and 24 children is correct!!Bravooo

8 men $8*4 =$32

32 children 32/4 = 8 groups of 4 = $8

and where are the women?

Hi dear friends!

Fourty poeple( men,women,and children)entered a resturant.

each man should pay 4$.

each woman 2$.

each 4children 1$.

totally they paid 40$.

howmany men were out of 40?

howmany women?

and howmany children?

have a nice time......

[Hats on a death row!! One of my favorites puzzles!]

Hello....I didn`t read all the answers.but if we concider the colors as 0 and 1

now let us take any random possibilty...e.g

00111000111001110011

the first one(0)has 50% survival chance,,but he will say the color seeing in the front of him,,,in this case (0)

the second one (he knew now his color),at the same time he saw the color in front of him (1),which is different color

so he will wait shortly then he`ll say his color(0).

the 3rd(is sure now that he has (1),because the 2nd didn`t said immediatly,,0).

and as the person infront of him has also (1)in this example,so he will say his color(1) immediatly!

and so on!

so...

1st....(o)

2nd.............(0)

3rd....(1)

4th....(1)

5th.............(1)

6th....(0)