Using Only Five 5's and any mathematical operator make sum as 37

If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.

I am the beginning of the end, and the end of time and space. I am essential to creation, and I surround every place. What am I?

Can you place three balls such that the equation shown in the picture holds true?

Decode below three cipher riddles:

26 L of the A
7 D of the W
1000 Y in a M

How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?

The more of this there is, the less you see. What is it?

What does below Rebus say?

XLR8

What is the next number?

77, 49, 36, 18 ?

What do you get if you add 2 to 100 four times?