Puzzle : The Case of Missing Servant

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. 

For example, servant 14 comes in and says "Servant 14, reporting in." 

One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one.

Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. 

Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. 

The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him.

How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?

Mathematical Trick to know the missing servant! 

Solution : The Missing Servant in the Case

What was the case?

When the first servant comes in, the king should write his number on the small piece of paper. For every next servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper while erasing old one.

Addition of numbers from 1 to 100 = 5050.

Hence, 

Missing Servant Number = 5050 -  Addition of ranks of 99 Servants.

So, depending on how far the addition of 99 servants' rank goes to near 5050, the king can easily deduce the rank of missing servant.

For example, if the addition that king has after 99 servants report in is 5000 then the servant having rank = 5050 - 5000 = 50 must be missing. 

 

Puzzle by The Mathemagician

Mavis the 'mathemagician' held ten cards (face down) in her hand - Ace (1) to 10 of Hearts.
She moved the top card to the bottom of the pack, counting '1', and turned up the next card, placing it on the table. It was the Ace.

She counted two more cards to the bottom of the pack, showed the next card - the '2' - and placed it on the table. She moved those counted top 2 cards to bottom.

Counting, 'One, two, three' more to the bottom, she then showed the next card - '3' of Hearts followed by moving those counted top 3 to bottom. 

This continued for four to nine, and the final card was - ta-daah! - the '10' of Hearts.

Question: What was the original order of cards, from the top to bottom?

THIS should be the order!

Solution of Mathemagician's Puzzle

What was the MAGIC?

Let us name ten cards as C1, C2, C3.....C10 and initially they are in order like below.

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

1. Mavis the 'mathemagician' moved the top card to the bottom of the pack, counting '1', and turned up the next card, placing it on the table. It was the Ace. 

Therefore, card C2 must be the Ace. C2 = ACE. 

So, the order has to be as -

C1 ACE C3 C4 C5 C6 C7 C8 C9 C10.

She moved top card to the bottom & kept Ace card on the table.

C3 C4 C5 C6 C7 C8 C9 C10 C1 

2. She counted two more cards to the bottom of the pack, showed the next card - the '2' - and placed it on the table. 

Therefore, the card C5 must be '2'. C5 = 2.

C3 C4 2 C6 C7 C8 C9 C10 C1 

Moved C3, C4 to the bottom while keeping C5 = 2 on the table.

C6 C7 C8 C9 C10 C1 C3 C4

3. Counted 3 more cards to the bottom of the pack, found '3' as next card. So, the card C9 must be '3'. C9 = 3.

C6 C7 C8 3 C10 C1 C3 C4

Moved C6, C7 and C8 to the bottom while keeping C9 = 3 on table.

C10 C1 C3 C4 C6 C7 C8

4. Counted 4 more cards to the bottom of the pack, found '4' as next card. So, the card C6 must be '4'. C6 = 4.

C10 C1 C3 C4 4 C7 C8

Moved C10, C1, C3 and C4 to the bottom while keeping C6 = 4 on table.

C7 C8 C10 C1 C3 C4

5. Counted 5 more cards to the bottom of the pack, found '5' as next card. So, the card C4 must be '5'. C4 = 5.

C7 C8 C10 C1 C3 5

Moved C7, C8, C10, C1, and C3 to the bottom while keeping C4 = 5 on table.

C7 C8 C10 C1 C3

6. Counted 6 more cards to the bottom of the pack where count goes to the top of the pack after 5, found '6' as next card. So, the card C8 must be '6'. C8 = 6.

C7 6 C10 C1 C3 

Moved C7 to the bottom while keeping C8 = 6 on table.

C10 C1 C3 C7

7. Counted 7 more cards to the bottom of the pack where count goes back to the top of the pack after 4, found '7' as next card. So, the card C7 must be '7'. .C7 = 7

C10 C1 C3 7

Moved C10, C1 and C3 to the bottom while keeping C8 = 6 on table. 

C10 C1 C3.

8.  Counted 8 more cards to the bottom of the pack where count goes back to the top of the pack after 3 and 6, found '8' as next card. So, the card C3 must be '8'. .C3 = 8

C10 C1 8

Moved C10, C1 to the bottom of the pack while keeping C3 = 8 on table.

C10 C1 

9. Counted 9 more cards to the bottom of the pack where count goes back to the top of the pack after 2, 4, 6 and 8, found '9' as next card. So, the card C1 must be '9'. .C1 = 9

C10 9. 

Keeping C1 = 9 on the table leaves only 1 card in the deck.

C10

10. The final card was - ta-daah! - the '10' of Hearts. Hence, .C10 = 10

So the initial order of

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

must be 

9 A 8 5 2 4 7 6 3 10