The sum of all the four numbers is 31.
Only one of number is odd.
The highest number minus the lowest number is 7.
If you subtract the middle two numbers, it equals two.
There are no duplicate numbers.
Jamie looked at his reflection on the window mirror of the 45th floor. Driven by an irrational impulse, he made a leap through the window on the other side. Yet Jamie did not encounter even a single bruise.
How can this be possible if he did neither landed on a soft surface nor used a parachute?
You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.
The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.
You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.
You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.
What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?
10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.
Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.
Here is how you would find one poisoned bottle out of eight total bottles of wine.
Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.
With ten people there are 1024 unique combination so you could test up to 1024 bottles of wine.
Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each.
Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die.
One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.
In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.
In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.
If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?
At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.
Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.
All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.
A strange tradition is followed in an orthodox and undeveloped village. The chief of the village collects taxes from all the males of the village yearly. But it is the method of taking taxes that is interesting.
The taxes paid in the form of grains and every male should pay equal pounds corresponding to his age. In simpler terms, a man aged 10 years will have to pay 10 pounds of grains and a 20 years old will be paying 20 pounds of grain.
The chief stands on a riser containing 7 different weights next to a large 2 pan scale. As per the interesting custom, the chief can only weigh using three of the seven weights.
In such a scenario, can you calculate what must be the weights of the seven weights each and who is the oldest man the chief can measure using those weights?
Jonathan has three boxes containing milk chocolates and dark chocolates. The problem is that all of them have been labeled incorrectly as follows.
Box1: Dark Chocolates
Box2: Milk Chocolates
Box3: Dark Chocolates and Milk Chocolates
How will he label all the boxes correctly by just opening one box?
It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.
We know that Christanio Ronaldo tells the truth on only a single day of the week. If the statement on day 1 is untrue, this means that he tells the truth on Monday or Tuesday. If the statement on day 3 is untrue, this means that he tells the truth on Wednesday or Friday. Since Christanio Ronaldo tells the truth on only one day, these statements cannot both be untrue. So, exactly one of these statements must be true, and the statement on day 2 must be untrue.
Assume that the statement on day 1 is true. Then the statement on day 3 must be untrue, from which follows that Christanio Ronaldo tells the truth on Wednesday or Friday. So, day 1 is a Wednesday or a Friday. Therefore, day 2 is a Thursday or a Saturday. However, this would imply that the statement on day 2 is true, which is impossible. From this we can conclude that the statement on day 1 must be untrue.
This means that Christanio Ronaldo told the truth on day 3 and that this day is a Monday or a Tuesday. So day 2 is a Sunday or a Monday. Because the statement on day 2 must be untrue, we can conclude that day 2 is a Monday.
So day 3 is a Tuesday. Therefore, the day on which Christanio Ronaldo tells the truth is Tuesday.
Once there lived a king who did not allow anybody to leave the kingdom and any foreigners in his kingdom. There was only one bridge that connected his empire with the outer world. A guard who was a sharpshooter was specially assigned for a lookout on the bridge. According to the orders, anyone moving outside should be killed and anyone coming to his kingdom should be sent back. To take rest, the guard used to sit inside his hut for 5 minutes and return back on the lookout. The bridge took a minimum of 8 minutes to pass.
Even then, a woman was able to escape the kingdom without incurring any kind of harm to the guard.
The woman started walking across the bridge when the guard was inside the hut. She walked all the time he was inside (5 minutes) and then turned and moved back towards the kingdom. On approaching the kingdom he was asked for papers and since she did not have any, she was sent back.
Four friends need to cross a dangerous bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.
Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins
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