What came first, the chicken or the egg?
Dinosaurs laid eggs long before there were chickens
You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.
Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.
So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.
There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3
(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.
That was the easy part.
What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.
Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.
Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.
For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.
Sherlock, A detective who was mere days from cracking an international smuggling ring has suddenly gone missing. While inspecting his last-known location, you find a note:
710 57735 34 5508 51 7718
Currently, there are 3 suspects: Bill, John, and Todd. Can you break the detective's code and find the criminal's name?
Bill. If you read the message upside down, you'll notice that the numbers resemble letters and that those letters form legible sentences. The message is 'Bill is boss. He sells oil.'
Can you decipher these quotes by Albert Einstein?
Blf xzm mvevi hloev z kilyovn lm gsv ovevo lm dsrxs rg dzh xivzgvw.
You can never solve a problem on the level on which it was created.
Here is the answer key:
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:
There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.
So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.
There is one four-digit whole number n, such that the last four digits of n2 are in fact the original number n
Looking at the last digit, the last digit must be either 0, 1, 5 or 6.
Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76.
Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376.
Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376.
Out of those, only 9376 is a 4 digit number