11(am) plus 2(hours) = 1(O’clock)

A and B cross first using up 2 minutes.
A comes back making it 3
C and D cross making it 13 minutes
then B crosses back over making it 15 minutes.
And finally A and B cross together to make it 17 minutes!

Cheryl’s birthday is on July 16

From the first statement, any of the day’s in that month should not be unique, else otherwise Albert can’t be sure of Bernard not knowing the birthday.Lets take the possible answers

If it was May => Albert cant say the first statement, as if the day was 19, Bernard will know the birthday for sure.
If, it is was June => Same is the case with 18th.
If, it is in July => Albert can say the first statement.
If it is in August => Albert can say the first statement.

Now after hearing first statement, Bernard can clearly figure out that the month of birthday is either July or August.

So, if the day was 14 => It can be July 14 or August 14 =>as bernard know now, it is not possible
If the day was 15 => It can be Aug 15
If the day was 16 => it can be July 16
If the day was 17 => it can be Aug 17

Only three days possible are Aug 15, July 16 or Aug 17, bernard knows the day so he know the birthday.

From the third statement, Albert knows the birthday as well, so it can only be July 16, as if the month was Aug, Albert can not figure it out.

Thus the Cheryl’s birthday is on July 16.

Answer – Only perfect square doors will be open at the end.

For example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8…) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..

question: what state are the doors in after the last pass? which are open which are closed?

solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9… leaving it open at the end. only perfect square doors will be open at the end.