Puzzle : Draw The Maximum Sum

Assume you are blindfolded and placed in front of a large bowl containing currency in $50, $20, $10, and $5 denominations. You are allowed to reach in and remove bills, one bill at a time. The drawing stops as soon as you have selected four-of-a-kind— four bills of the same denomination. 

What is the maximum sum of money you could accumulate before the drawing ends?

THIS should be the maximum sum! 

Source

The Maximum That You Can Get!

What was the question?

The quick response to the question by any body would be 3 x 50 + 1 x 20 = $170 would be the maximum as next pick of highest currency of $50 will stop drawing of currencies. But that 5th attempt may not be of $50 & could be of $20,$10 or $5. So this case doesn't count those possibilities. This is worst case scenario where 3 out of 4 picks resulted into accumulation of 3 currencies of same denomination of $50.

So ideally, after drawing three $50 s, three $20 s, three $10 s, three $5 s and finally drawing one anything will stop drawing attempts as that will accumulate 4 bills of same denomination (of $50/$20/$10/$5). The best that can be picked is $50 to get the maximum.

That is total maximum of 50 x 3 + 20 x 3 + 10 x 3 + 5 x 3 + 50 = $305 can be accumulated in 12 attempts & stopping after 13th attempt. And this will be the best case.

Puzzle: A Visit to the Casino!

There is a casino and it has 4 gates, let say A, B, C and D. 

Now the condition is that every time you enter casino you have to pay $5 and every time you leave the casino, you again have to pay $5. Also, whenever you enter the casino whatever amount you have with you will get double.

Now you enter the casino through gate A and come out through gate B, again you go inside casino from gate C and come out of gate D, at the end of this process you should be left with no money? 

So calculate how much money you should carry with you when you enter the Casino?

THIS should be the amount that you need to carry... 

Solution: Amount Needed for a Casino Visit

What is the puzzle?

Let's suppose you have amount 'x' initially in your wallet.

1. On paying $5 for entry at gate A amount left is x - 5 .

2. After entry into the casino it double and becomes 2 ( x - 5 ) = 2x - 10.

3. For exit at gate B you pay $5 and the amount left with you is 2x - 10 - 5 = 2x - 15.

4. Again, for the entry at gate C, you pay $5 more. So the amount with you will be  - 
    2x - 15 - 5 = 2x - 20.

5. The amount is doubled to become 2(2x - 20) = 4x - 40 after the entry into casino.

6. Now, you have to pay $5 one more time to have exit via gate D. Hence, the amount 
left will be 4x - 40 - 5 = 4x - 45.

7. As per given condition, the amount that you should have on exit must be 0.

Hence, 4x - 45 = 0 i.e. 4x = 45. Therefore, x = 11.25. 

So you should carry $11.25 before you enter into the casino to have $0 after exit out of the casino.

Let's verify this one with the amount at each of stages above.

1. Amount = 11.25 - 5 = 6.25

2. Amount = 2 x 6.25 = 12.50

3. Amount = 12.50 - 5 = 7.50

4. Amount = 7.50 - 5 = 2.50

5. Amount = 2 x 2.50 = 5 

6. Amount = 5 - 5 = 0

7. Amount = 0
    

Maze Challenge For a Rat?

A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.

What is the expected amount of time it will take for the rat to finish the maze?

This could be the average time that rat needed!
 

A Rat Finishing off The Maze!

The challenge ahead of rat?

For rat, there are 2 paths viz A (5 minutes) and C (3 minutes) leading to the end while paths B (8 minutes) and D (2 minutes) lead to the start again.

Since, there are 4 paths & each having equal chance of being chosen by rat, there is 1/4 th chance for each path for to be chosen by rat.

Let's assume T be the time needed for rat to finish the maze. 

But if rat selects path B or D then rat need T more time again as these paths lead to the start of the maze again.

Hence,

T = (1/4) x A + (1/4) x B + (1/4) x C + (1/4) x D

T =  (1/4) x 5 + (1/4) x (8 + T) + (1/4) x 3 + (1/4) x (2 + T)

T = (5/4) + (2) + (T/4) + (3/4) + (1/2) + (T/4)

T =  (9/2) + (T/2)

T/2 = 9/2

T = 9

That is rat needs 9 minutes to finish the maze. 

 

Challenge of Father to Son

A man told his son that he would give him $1000 if he could accomplish the following task. 

The father gave his son ten envelopes and a thousand dollars, all in one dollar bills. He told his son, "Place the money in the envelopes in such a manner that no matter what number of dollars I ask for, you can give me one or more of the envelopes, containing the exact amount I asked for without having to open any of the envelopes. If you can do this, you will keep the $1000."

When the father asked for a sum of money, the son was able to give him envelopes containing the exact amount of money asked for. 

How did the son distribute the money among the ten envelopes?

THIS is how son accepts the challenge!

Son's Response to the Father's Challenge

What was the challenge?

For a moment, let's suppose father had given $30 to son and provided 5 envelopes and put the same challenge.
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Here, use of the binary number system helps in matter.

The son would distribute 15 dollars into 4 envelops like - 

2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8.

Now, for any amount asked between 1 to 15, son can produce some of these 4 envelops wherever 1 is there in envelop column for that particular amount.

For example, if father asks for $10 (Binary - 1010), son would give envelop 4 and 2 (8+2=10).

After putting $15 dollar in 4 envelops, he puts remaining $15 in 5th envelop so that he can cover rest of amount between 16 to 30.

If father asks amount greater than 15 then he would take envelop of $15 first and depending on how much the amount asked is greater than this $15 he would pick some of those 4 envelops.

For example, if father asks for $24, then he picks envelop 5 having $15 and envelops for amount 24 - 15 = 9 (Binary - 1009) i.e. envelop 4 and 1 (8+1=9)  i.e. total of 15 + 9 = 24.

So, what we observe from this is that the number of envelops needed for such arrangement is equal to the number of binary bits needed to represent the amount itself or nearest power of 2 greater than the amount.

In above case, to represent 30 in binary we need 5 bits or nearest power of 2 greater than 30 is 32 which needs 5 bits for representation.

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Now, let's turn to the actual challenge where father has asked son to distribute $1000 rightly in 10 envelopes. 

The reason for selecting 10 as a number of envelops is clear now as 1000 needs 10 bits in binary or nearest power of 2 greater than 1000 is 1024 which needs 10 bits for binary representation.

So, the son puts 256, 128, 64, 32, 16, 8, 4, 2, 1 dollars in 9 envelops (envelop numbered as Envelop 9, Envelop 8.........Envelop 1 in order)
 and 1000 - 511 = 489 dollars in 10th envelop.

First 9 envelops will cover amounts from 1 to 511 and for amounts greater than 511 inclusion of 10th envelop having 489 dollars is mandatory.

Again selection of envelops for the amount 511 to 1000 depends on how much the amount exceeds the $489. The binary representation of that difference and selection of envelop accordingly is all that needed.
 
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Let's make sure this distribution with couple of examples.

If father asks for amount of $109 (binary - 1101101) then son picks 

Envelop 1($1) + Envelop 3 ($4) + Envelop 4 ($8) + Envelop 6 ($32) + 
Envelop 7 ($64) i.e. having amount = 1 + 4 + 8 + 32 + 64 = 109 dollars.

If father asks for $525 then son gives $489 via Envelop 10 and rest of amount 
530 - 489 = 40 (Binary - 101000) in form of Envelop - 6 ($32) and Envelop 4 ($8).   

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Generous Devotee

A devotee visits 9 temples when he visits India. All these nine temples have one thing in common - there are 100 steps in every temple. The devotee puts Re.1 coin after climbing up every step. He does the same while climbing down every step. At each temple, the devotee offers half of his money from his pocket to god. In this way, his pocket becomes empty after his visit to 9th temple.

Can you calculate the total amount he had initially ? 

Click here to know exact amount! 

Donations By The Devotee

Why to calculate those?

Using algebraic equations in the case can make things complicated unnecessarily. Hence, we would start from backward. Before putting 100 coins on steps while climbing down 9th temple devotee must had 100 coins. That means he had 200 coins when he climbed up the 9th temple half of which i.e. 100 he offered to that temple & 100 put on the 100 steps of 9th temple. Moreover, he must have placed 100 coins while climbing up 9th temple. So before visit to 9th temple he must had, (100 x 2) + 100 = 300 coins.

Same way, finding the amount he had before visit to each temple like below.

Before eight temple: (300+100)*2 + 100 = 900
Before seventh temple: (900+100)*2 + 100 = 2100
Before Sixth temple: (2100+100)*2 + 100 = 4300
Before fifth temple: (4300+100)*2 + 100 = 8900
Before fourth temple: (8900+100)*2 + 100 = 18100
Before third temple: (18100+100)*2 + 100 = 36,500
Before second temple: (36500+100)*2 + 100 = 73300
Before first temple: (73300+100)*2 + 100 = 146900

 
To conclude,  he had Rs. 146900 initially.