Solution:

1 2 3 4 5 6 and 7 8 6

Since summation of numbers 1-9(1 2 3 4 5 6 7 8 9) is 45 , so its impossible to divide into two equal parts.
Since we cant flip over 6 to make it 9 but we can flip 9 to make it 6 and hence the solution

Solution:

The women is a of short stature. She can't reach the upper elevator buttons, but can push is with his umbrella.

Solution:

2 people

Say 2 helpers are Helper1 & Helper2

Helper1 food Helper2 food McDuck's Food
End of Day 1 : 1 4 4
Now 'A' will head back

End of Day 2 : N/A 2 4
Now 'B' will head back

Now Mr. McDuck have 4 day food :-)

Solution:

Fifteen Minutes.

After frying for five minutes, you can take out one steak and put the third one inside while turning the other one inside. After five more minutes, one steak is fried from both sides and you can take it out. Now put in the steak that we removed and turn the other one inside. After five more minutes, they both will also be fried from both sides.

Solution:

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.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

There is a simple logic to solve this question.
The size of the chocolate is 2 x 8. Thus, you need to have 2 x 8 = 16 pieces.
Every time you break the chocolate, you will get one extra piece.
Thus, to get 16 pieces, you must break it (16 - 1) = 15 times.

Solution:

Solution is below

Solution:

7 Races.

Solution:

Sansa has the siamese (conjoined) twin sister.

Solution:

17 min

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)

Total time = 2 + 2 + 10 + 1 + 2 = 17 minutes