Can you checkmate all the kings in less than 10 moves?
Following are the rules:
1. White can make up to 10 legitimate moves until all kings are checkmate.
2. You can take any black piece(s) except the King.
3. Your king cannot be in check position at any time of the game.
You can see the figure or draw one of your own. The scenario is as shown. There are three houses represented with the triangle over the square. There are three utilities: W, G and E representing water, gas and electricity respectively.
Can you draw a line and get each utility into every house (9) total lines without ever crossing any line?
We all know that New Year occurs after a week from Christmas and thus it falls on the same day as of Christmas. But this will not happen in 2050. In 2050, Christmas will appear on Sunday while New Year will appear on Saturday.
Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
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 minutes. 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)
A man desired to get into his work building, however he had forgotten his code.
However, he did recollect five pieces of information
-> Sum of 5th number and 3rd number is 14.
-> Difference of 4th and 2nd number is 1.
-> The 1st number is one less than twice the 2nd number.
->The 2nd number and the 3rd number equals 10.
->The sum of all digits is 30.
We have arranged matchsticks in a manner that all the first row, second row, first column and third column contain 12 matchsticks each. Can you remove 4 matchsticks and rearrange all the remaining matchsticks that we are still left with 12 matchsticks in the first row, second row, first column and the third column.
A homicide team enters a crime scene where a dead body of a fat old man lies there on the floor with blood oozing out of his head. The victim is holding a gun and a tape recorder lies there by his side. One of the detective picks up the recorder and plays it "I am tired of this life and hence I have decided to relieve myself from the worldly pains". A gunshot follows the message.
The teams starts investigating a murder investigation.
A mastermind organized a quiz competition in which six selected candidates were invite namely James Hunt, Ruxandra Bar, Sophia Connors, David Finch, Fred Odea and Brian Miller. A 'special puzzle' was asked to all of them. The first one to answer it was promised for a big award.
After that, the candidates were offered the meal before the mastermind stood up to announce the much awaited result. He started announcing:
'Ok now everybody!'
'The winner of…..'
'The Hardest Riddle Ever Event.'
And then he smiled. All the candidates understood who won.
Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.
Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'