Puzzle 102 - fake coins

-

 


Thirty €1 coins are arranged in a circle on a table. Ten of these coins are fake and a little lighter than the 20 real coins. You don’t know where in the circle the fake coins are, but you have been told they are all directly next to one another in the circle. Your task is to find as many real coins as possible. To do this, you have at your disposal a beam balance, a device that can be used to compare the masses of two objects or sets of objects. Your beam balance does not have weights, and it can be used only once. Which coins should you weigh, and how many real coins can you definitely find?

You can find the solution here.

Comments

Post a Comment

Popular posts from this blog

Puzzle 9 : The 4 coins problem

-
 Solution to puzzle 8: Maria is not counting Sundays. So she is not counting 1/7th of days of her age.  That is to say she is only counting 6/7th of her age. 6/7th of her age = 30 Her actual age = 30 X 7/6 = 210/6 = 35 years.  Puzzle 9 : The 4 coins problem  You’re creating a new coin system for your country. You must use only four coin values and you must be able to create the values 1 through 10 using one coin at a minimum and two coins maximum. What 4 coins do you choose, and can you think of a second set of 4 coins that achieves the same goal?
Read more

Puzzle 35 - Open sesame

-
I am writing a post after two months.  Here is a locked question ;) There is a number lock. You have to find the number to unlock it.  And here are the results of various number combinations. 682 - one digit is right and in the right position. 614 - one digit is right but in the wrong position. 206 - two digits are right but both are in the wrong positions. 738 - all digits are wrong. 380 - one digit is right but in the wrong place So what is the correct code to open the lock? Find the solution here
Read more

Puzzle 66 - solution

-
 The question was  What is wrong with this proof ? a=b multiply both the sides by a     --1 a2 = ab  add -b2 to both the side           --2 a2-b2 =ab-b2  factorizing                                --3 (a+b)(a-b) = b(a-b)  divide both the sides by (a-b)    ---4 a+b=b Substituting a with b,              ---5 2b = b Dividing both the sides by b    ---6 2 = 1   The problem is with step 4. When a is equal to b, a-b is zero. You can not divide numbers by zero because division is only defined for non-zero numbers.  
Read more