Hard Riddles

We can calculate out how many days there will be for the 13th on each month if we count from the beginning of the year (January 1). Then we divide total days by 7 to get the remainders. We also need to consider the leap year. Through the whole year we had all kinds of remainders, from 0 to 6. The minimum of occurence for all the unique remainders was 1. It means that we have at least one Friday on 13th. In a regular year, the best chance you can get 3 Fridays on 13th, which are in February, March and December because the remainders of these 3 months are 2. In a leap year, the best chance you also can get 3 Fridays on 13th, which are in January, April and July because the remainders of these 3 months are 6.

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.

As per the puzzle given ablove The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refuelled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.

What goes up must come down.

1:45.

The man gave away a total of 25 cents. He divided it between two people. Therefore, he gave a quarter to two

6526884. To get from 4 to 12, you use the following formula. 12 = (4 squared)/2 + 4.

Continue on with (X^2)/2 + X.

he gradient of the teal hypotenuse is different than the gradient of the red hypotenuse.

Wise you are, wise you be, I see you are too wise for: ME!

No time at all it is already built

1) 8/(3-(8/3))
= 8/(1/3)
= 24

2)((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24

3)(3!/)*8

4)(8-3)!/(8-3)
( × )!
( + )!
√(8×8×3×3)
8+(8×(3!/3))
((√(8+8) × (3/3))!
√(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!