Solution:

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.

Solution:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

Solution:

40 balls

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

A Question

Solution:

19

Explanation:
The children can purchase 19 chocolates in return.

Out of 77 wrappers, 75 will be used to buy 15 chocolates and two will be left spare.
The 15 chocolates will create 15 empty wrappers that can be exchanged to get three chocolates.
Three chocolates will return three wrappers which will help them buy another chocolate.
Now the wrapper from this chocolate and the two spare that were left earlier will get them another chocolate.
15 + 3 + 1 = 19

Solution:

192 021

If you look at it as two-digit numbers you'll see:10 11 12 13 14 15 16 17 18 19 20 21

Solution:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa

Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

Solution:

They used the planks in the manner that is shown in the figure.

Solution:

1/36

Explanation:
The number of possibilities for the first as well as the second dice are 6 and 6 respectively. Therefore the total possibilities or outcome are 6 * 6 = 36.
Out of 36 outcomes, we need only one case i.e. the first gives 2 and the second gives 5.
Therefore, the probability will be 1 on 36 or 1/36.