Let us break the problem into different parts.
We know that after six, all the numbers that are divisible by three can be ordered. This is because we can expand them as the sum off sixes and nines.
After twenty six, the numbers that are divisible by three if subtracted by twenty can also be ordered.
After forty six, all the numbers will fit in one of the already discussed category. Thus all the numbers can be ordered.
So forty three is the last number that does not come under any of the already discussed category.
44 = 20 + 6 * 4, 45 = 6 * 6 + 9
A teacher is told that the principal of the school will be inspecting his class on the next day. Now, the teacher is worried for the impression that his class might cast on the principal since all the students are not intelligent. Also, the principal can ask questions from anywhere. However, he will have the power to choose any student for answering the question.
Now he wants that the principal must be impressed with the performance of his class. What will he do to maximize the final impression on the principal ?
The teacher will use a simple trick to form a perfect impression. He will ask all his students to raise their hands on each of the question that is asked. But the only catch will be that those who knows the answer correctly will raise their right hand and others will raise their left hand.
In this way, the principal will see all the hands being raised for each question even though all won't be knowing the correct answer. The teacher will ask only those who know the answer and they will always be correct. So the principal will be impressed to full extent.
A thief was running from the police after the biggest theft the town saw. He took his guard in one of the thirteen caves arranged in a circle. Each day, the thief moves either to the adjacent cave or stay in the same cave. Two cops goes there daily and have enough time to enter any two of the caves out of them.
How will the cop make sure to catch the thief in minimum number of days and what are the minimum number of days?
Your friends are coming over to your birthday party. Your mum has bought this delicious giant doughnut for you guys. Unfortunately she had to go somewhere. Now you have to serve all of them including you.
There are a total of nine kids including you. None of you would mind a smaller piece. Can you cut this doughnut into nine pieces in just three straight cuts?
You have ten boxes and an electronic weighing machine. In those ten boxes, you have chocolates. Each chocolate weigh 20 grams. But in one box the chocolates are defective and each weigh 19 grams exactly.
Now you can weigh in the electronic weighing machine but you can use that machine just once. How will you find out which box has the defected chocolates.
If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.
Let us begin by labelling boxes as 1, 2, 3 and so on till 10.
Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)
The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.
You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.
John is out with his class of 25 boys to a local park. Each guy has a remote controlled car with them. The park has a racetrack that allows 5 cars to be raced at once. Their teacher, Mr. Ted, declares that the top three fastest cars get ice cream.
How many races are required to determine the 3 fastest cars?
A professor gives a set of three questions to the most brilliant students of his university. You can see the questions in the attached image if required. To his surprise, there are different answers by all three of them. Below are the answers by them:
Now you have the information that each one of them has given one answer wrong, can you find out the real answers to every problem?
Since each one of them gave one answer wrong, this means that each one of them gave two answers right.
Let us assume that Student A gave a wrong answer to the first question. This will mean that Student B also gave a wrong answer for the first. This will conclude that the rest of the two answers given by them are correct. However, the answers are different and thus it is not possible.
Thus both Student A and Student B must be right with the first question and the answer to the first is two.
If you keep applying the same logic, you will come to a conclusion that following are the correct answers:
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