Solution:

I am word "Introvert"

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

Solution:

100

Series of number that do not contain albhabet 'T'

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.

Solution:

Selfie
Explanation:
: Selfie = is six letter word
: Self = The first four letter is me.
: e = The second and last letter are the same.
: fee = The fourth second and last letter are payment.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

16

Explanation:
9 [1x1] squares
5 [2x2] squares
1 [4x4] square
1 [4x4] square

9 + 5 + 1 + 1 = 16 squares.

Solution:

2 cakes

At each toll bridge you are required to give half of your cakes, and you receive one back. Which leaves you with 2 cakes after every bridge.

Solution:

Batsman hit six and gone retired hurt while the new batsman come to play and get runout for his wicket and the other batsman hit six and made both century and team win..

Solution:

Step one: First you will have to make 3 trips of 1,000 apples 333 miles. Then you will be left with 2,001 apples and 667 miles to go.

Step two: Next you will have to take 2 trips of 1,000 apples 500 miles. In this manner you will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind).

Step three: At last, you will have to travel the last 167 miles with one load of 1,000 apples and you will be left with 833 apples when you reach the town