Solution:

It can be done as shown in the picture below. Key to solving such problems is to fill the prime number first.

Solution:

16-days

In 15-days, he will climb 15-inch and when he climbs 5 inch more, he will be out of the well :-)

Solution:

Firstly: Policeman should take the thug across the river thames and escort the boat back with him. (Thug across, everyone else on this side)
Then, Policeman will take one cat along, leave it there, and on the way back, bring back the thug (since thug cannot be left with anyone without policeman's supervision).
Then Mrs. Mother will take the second cat on the other side. Drop it there and bring the boat back. For the 4th trip, Mrs. Mother And Mr. Father Will travel to the other side, and Mrs. Mother will stay there, and MR. Father will bring the boat back. (Now Mrs. Mother and 2 cats are on the other side.)
5th trip, the policeman takes the thug across. Both of them stay there and Mrs. Mother will come back. Then for the next trip, Mr. Father and Mrs. Mother will go on the boat, and Mrs. Mother will stay there, and Mr. Father will return back.
Then for the 7th trip, Mr. Father will travel with one dog. On the return, the policeman and the thug will return. Then in the next trip, the Policeman will take the second dog. (Now Mrs. Mother and her two cats; Mr. Father and his two dogs and the Policeman is here).
Now finally the policeman takes the boat on the other side and brings back the thug.
Tough one, right ?

Solution:

20
Just count the nines in the numbers: 6, 16, 26, 36, 46, 56, 76, 86, 96, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
Note: 66 contains TWO nines

Solution:

13 x 3 + 1 = 40

The number of decks is absolutely not relevant here which means whether we mix 5 or 500 cards still results would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
If you are lucky just 5 cards of the same kind can be obtained in 4 steps

The unluckiest(worst) way is our solution as we need to guarantee a four of a kind.
Draw 3 of each kind =>now we have 39 cards. Next card will guarantee 4 of a kind.

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

The probability of two numbers being relatively prime is 60%. Thus if you keep playing the bet, you will hypothetically make 20 cents on every play.

Thus, you should accept the bet.

Solution:

It can be done as shown in the picture below.

Solution:

At 01:05

Solution:

The check was for dollars 31.63.

The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.

Let x be the dollars of the check, and y be the cent.

The check was for 100x + y cent
He was given 100y + x cent

Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5

or 199x ≡ -5 (mod 98)
or 98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
this quickly leads to x = 31

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