Solve the alphametic puzzle by replacing the alphabet with the number.

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Queen Elizabeth organised a royal party.

To Avoid uninvited guest royal family set a password.
Jack (an uninvited person) plan to enter the party. He stand nearby the door.

First guest comes, the security person said 'twelve' and guest replied with six.
Second guest comes , the security person said 'six' and guest replied with 'three'.

Jack thought is enough and he walked to the entry point. The security person said 'eight' , Jack replied smilingly 'four'.
He was immediately thrown out of the party. why ?

Find the next number in the series.
1 10 24 43 67 ?

Place the numbers from 1 to 9 in the circles in a manner that each side of the triangle sums up to 17.

Below the four parts have been reorganised. The four partitions are exactly the same in both arrangements. Why is there a hole?

A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?

Which word is least like the others? Third, fourth, fifth, sixth, seventh, eighth, ninth?

I am a mother. However, I never give birth.

I am a father. However, I never nurse my children.

Wandering is not possible for me, but standing still happens occurs rarely.

Who am I?

A cat, a dog and a monkey were stolen. 3 suspects got caught: Harish, Manoj and Tarun. All we know is each person stole one animal, but we do not know who stole which. Here are the investigation statements. Harish said: Tarun stole the cat. Manoj said: Tarun stole the dog. Tarun said: They both were lying. I did not steal the cat or the dog. Later on, the police found out the man who stole the monkey told a lie. The man who stole the cat told the truth. Can you find out who stole which?

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.