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.

81 x 9 = 801. What must you do to make the this equation true?

There is a shop where written:
Buy 1 for $1
10 for $2
100 for $3
I needed 999 and still only paid $3. How could this be financially viable for the shop-keeper?

What can you deduce from the following BrainBat?

CRY
LMKI

What is the difference between a jeweller and a jailer?

Solve the following series:
AZ, GT, MN, __, YB

When is the top of a mountain similar to a savings account?

'S' refers to the number of seconds in a day and 'H' refers to the number of hours in ten years. Which quantity out of these two is greater?

what does this picture(rebus) riddle means ?

Railroad Crossing, look out for the cars. Can you spell that, without any R's ?