Solution:

10 Bulbs

For this tricky puzzle, you must check how many light bulbs in the row are having an odd number of factors. The first one surely has odd number of factors, the second has even, four has odd. Thus the bulb four and one will remain lit. The bulbs that are going to remain lit are perfect squares as they have an odd number of factors - 1, 4, 9, 16.

Since there are 100 passes, you can go up to 10 times 10 i.e. the square of 10. There are 10 perfect squares available to you - one, two, three, four, five, six, seven, eight, nine and ten. They corresponds to the bulb number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. All of them will remain lit and thus total ten bulbs will remain lit after 100 passes.

Solution:

Though the question is very simple, most of the people answer it wrong. The real time will be 9:30. If you don't believe us, take a clock with 9:30 time in front of the mirror and look for yourself.

Solution:

We agree that the series is quite difficult to crack and much tricky as well.

1 can be written as one (consist of 3 letters)
2 can be written as two (consist of 3 letters)
3 can be written as five (consist of 5 letters)
4 can be written as four (consist of 4 letters)
5 can be written as five (consist of 5 letters)
6 can be written as six (consist of 3 letters)

Thus the next term in the series is 3.

Solution:

If you are thinking of how to cut in such a way that you are able to make seven equal pieces, you are thinking in the wrong direction. You should give more stress on how to make the transaction with the worker.

You have to break the gold bar two times so you have the following sizes of the bar:

1/7 ---- Bar A
2/7 ---- Bar B
4/7 ---- Bar C

When the day 1 ends:
Give Bar A (You remain with Bar B and Bar C; Worker possess - Bar A)

When the day 2 ends:
Give Bar B and take the Bar A back (You remain with Bar A and Bar C; Worker possess - Bar B)

When the day 3 ends:
Give Bar A (You remain with Bar C; Worker possess - Bar A and Bar B)

When the day 4 ends:
Give Bar C and take back Bar A and Bar B (You remain with Bar A and Bar B; Worker possess - Bar C)

When the day 5 ends:
Give Bar A (You remain with Bar B; Worker possess - Bar A and Bar C)

When the day 6 ends:
Give Bar B and take back Bar A (You remain with Bar A; Worker possess - Bar B and Bar C)

When the day 7 ends:
Give Bar A (You remain with no bar; Worker possess - Bar A, Bar B and Bar C)

Thus by the end of seven days, you have taken care of everything.

Solution:

Side A = 3 Cannibals, 3 missionaries
Side B = 0 Cannibals, 0 missionaries

1 cannibal and 1 missionary goes in the boat
Side A = 2 Cannibals, 2 missionaries
Side B = 1 Cannibals, 1 missionaries

1 missionary returns back and 2 cannibals go there
Side A = 0 Cannibals, 3 missionaries
Side B = 3 Cannibals, 0 missionaries

1 cannibal returns and 2 missionaries go
Side A = 1 Cannibals, 1 missionaries
Side B = 2 Cannibals, 2 missionaries

1 missionary and 1 cannibal return and 2 missionaries go
Side A = 2 Cannibals, 0 missionaries
Side B = 1 Cannibals, 3 missionaries

One cannibal back, two cannibals go
Side A = 1 Cannibals, 0 missionaries
Side B = 2 Cannibals, 3 missionaries

One cannibal back, two cannibals go
Side A = 0 Cannibals, 0 missionaries
Side B = 3 Cannibals, 3 missionaries

Thus all have crossed successfully