Minimum number of cuts to the bar of gold that will allow you to pay to worker 1/7th each day ?

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?

Solution
Gold bar need minimum two cuts. It should be divided in powers of two.
e.g. for i=0 to k where 2^k is nearly equals to N length of gold piece. 1 ,2 , 4........assuming 7 parts which is divided into a total of 1,2,4.

first day : 1 ---(1)
second day: take back one and give 2----(2)
third day : give back one----(1,2)
fourth day : take back 2,1 and give 4 ----(4)
fifth day : give back 1------(1,4)
sixth day : give 2 and take back 1-----(4,2)
sevent day : give back one-------(1,2,4)

9 Digit Number Problem Every nth number formed by n digit is Divided By Every nth digit

Arrange the digits 0 to 9 such, that the number formed by the
first digit is divisible by 1, the number formed by the first two
digits is divisible by 2, that formed by the first three digits
divisible by 3, and so forth, thus the number formed by the first
9 digits will be divisible by 9 and that formed by all 10 digits
divisible by 10.

Here is what we would do.

_ _ _ _ _ _ _ _ _ _

those are my ten blanks.

Now let's start asking questions.

Since the whole thing has to be divisible by 10, the last number
must be a zero.

And since the first five digits must be divisible by 5, the fifth
digit must be a 5 or 0, but we have already used 0 so it must be a
five.

One last quickie is that the 2nd, 4th, 6th, and 8th and last
digits must be even, so all the rest are odd

So let's look at we have. I'll write the digits that could go in a
certain place under where they could go:


_ _ _ _ 5 _ _ _ _ 0
1 2 1 2 2 1 2 1
3 4 3 4 4 3 4 3
7 6 7 6 6 7 6 7
9 8 9 8 8 9 8 9

Now things get trickier.

Let's start with the 4th spot.

Any number is divisible by 4 if the last two digits are divisible
by 4

So what are the two-digit numbers divisible by 4 that begin with
an odd number that is not 5?

12 16 32 36 72 76 92 96

Hey, so the 4th digit is either a 2 or a 6.

Now, let's look at the 8th digit. If the first 8 digits are
divisible by 8, then they are divisible by 4 also. So if we apply
the same logic as above, we find that the 8th digit is either a 2 or
a 6.

So spaces 2 and 6 must be 4 or 8.

That narrows it down. So we have this:

_ _ _ _ 5 _ _ _ _ 0
1 4 1 2 4 1 2 1
3 8 3 6 8 3 6 3
7 7 7 7
9 9 9 9



Now for my next trick let's recall that every third space must be
divisible by 3. So every three numbers must sum to be divisible
by 3.

In particular spaces 4, 5, and 6 must sum to 3, and we already
have a five. So either we have

2 5 8 or 6 5 4

in positions 4, 5, and 6.

That is good stuff. That means we have either this:

_ 4 _ 2 5 8 _ 6 _ 0
1 1 1 1
3 3 3 3
7 7 7 7
9 9 9 9

or this:

_ 8 _ 6 5 4 _ 2 _ 0
1 1 1 1
3 3 3 3
7 7 7 7
9 9 9 9

Let's check each of the four digits we could put in the 8th place.
Since a number is divisible by 8 if its last 3 digits are, we need
number in the 6th, 7th, and 8th places to be divisible by 8. We see
that 816 and 896 work in the first case, and 432 and 472 work in the
second case. So we have:

_ 4 _ 2 5 8 _ 6 _ 0
1 1 1 1
3 3 9 3
7 7 7
9 9 9

or this:

_ 8 _ 6 5 4 _ 2 _ 0
1 1 3 1
3 3 7 3
7 7 7
9 9 9


Let's look at the 9th place now. Recall that 1+2+3+4+5+6+7+8+9 = 45,
so we don't need to worry about the 9th place - it will always work
out.

Let's look at the 1st and 3rd places. In case 1, only 147 and 741 are
options. In case 2, we could have 183, 189, 381, 387, 783, 789, 981, or
987. Note that 387 and 783 don't work, because then we don't have any
digits left for the 7th digit.

So our first case becomes:

_ 4 _ 2 5 8 9 6 3 0
1 1
7 7

We can just check the two numbers this can give rise to, namely 1472589630
and 7412589630. Since neither of them are divisible by 7, case one must be
a dead end. So we have to work with case two if we want a solution:

_ 8 _ 6 5 4 _ 2 _ 0
1 1 3 1
3 3 7 3
7 7 7
9 9 9

(and the first 3 digits are 183, 189, 381, 789, 981, or 987)

If we look at the 7th, 8th, and 9th digits to see if we can get them to sum
to three, we see that only 321, 327, 723, and 729 are options.

So which of our options for digits 1,2,3 and 7,8,9 can we use together? In
order to not repeat digits, we could use these:

183 and 729
189 and 327
189 and 723
381 and 729
789 and 321
981 and 327
981 and 723
987 and 321

This means that the only possible solutions to the puzzle are these:

1836547290
1896543270
1896547230
3816547290
7896543210
9816543270
9816547230
9876543210

This is now a small enough number of cases that we can check them by hand.
The first thing we should do is check the first 7 digits for divisibility by
seven. Here are the results:

1836547/7 = 262363.857142857...
1896543/7 = 270934.714285714...
1896547/7 = 270935.285714286...
3816547/7 = 545221
7896543/7 = 1128077.57142857...
9816543/7 = 1402363.28571429...
9816547/7 = 1402363.85714286...
9876543/7 = 1410934.71428571...

Monty Hall Problem

You are a contestant on a game show. You have three closed doors in front of you. One of the doors has a car behind it and the other two doors have nothing. You have to choose a door to open.

You have with you a Magic Watch.
- Whenever you ask the watch a question with a yes/no answer, it will blink either red or blue.
- One of the colors represents 'yes' and the other 'no'.
- You don't know which color means what. (gotcha!)
- You are allowed to ask the watch only 2 questions before you make your decision.

Which questions would you ask and how would you choose the door to open?

The puzzle might have a number of solutions and this is one of them.

Solution:

The contestant should ask the following two questions:
1) If I asked whether the car was behind door 1, would you blink red?
2) If I asked whether the car was behind door 2, would you blink red?

There are two cases here:
a) red = no, blue = yes:

The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.
b) red = yes, blue = no:

The watch can answer in 4 different ways:
1) red, blue:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2st question, car is not behind door 2.
2) blue, red:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is behind door 2.
3) blue, blue:
From the answer to the 1st question, car is not behind door 1.
From the answer to the 2nd question, car is not behind door 2.
4) red, red:
From the answer to the 1st question, car is behind door 1.
From the answer to the 2nd question, car is behind door 2.

Observe that the conclusions are exactly the same for the same answer set irrespective of the color code.
Hence, when the watch answers:
1) red, blue: Pick door 1.
2) blue, red: Pick door 2.
3) blue, blue: Pick door 3.
4) red, red: This outcome is impossible since it implies that the car is behind door 1 as well as door 2.

How can you get a fair coin toss

Interview question: How can you get a fair coin toss if someone hands you a coin that is weighted to come up heads more often than tails?

The story of these types of questions starts at the software giant Microsoft who pioneered the use of puzzle type question, where the aim is to test the interviewers problem solving skills and creativity. It's ironic that these type of questions are no longer asked in Microsoft job interviews, but the rest of the software industry hasn't caught on.



The solution is to toss the coin twice, which gives four possible outcomes:
H (heads) followed by H, probability of this is pH * pH
H followed by T (tails), probability of this is pH * pT
T followed by H, probability of this is pT * pH
T followed by H, probability of this is pT * pT
Looking at the above outcomes we can see that HT (H followed by T) is just as likely as TH,
So to get a fain coin toss, toss the coins and use HT or TH as fair outcomes of the toss, and discard the results of TT and HH.

Weigh System With More Explaination..How Many Weight We Needs..???

Puzzle: You have a common balance with two sides. You can only keep weighs on one side. You have to weigh items from 1 to 1000 weight units with integral weights only.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.

Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.

Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.

Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.

Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.


Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.
You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.
The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.
Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.

Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles
Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1

To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2

Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.

To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4

Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.

Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.

Pills Puzzle

A person was prescribed to take two pills (tablets), one each, from the two bottles viz. Bottle A and Bottle B, daily. The tablets are exactly look alike.

The medicines have to be taken exactly one tablet from each bottle, neither less nor more, else the medicines will not be effective.

One fine day, the patient popped out one tablet from Bottle A, but while taking the tablet from bottle B, by mistake, two tablets spilled over. Now he has three tablets in his hand, and he can't put back the extra tablet to Bottle B as all the tablets are identical in looks.

He has to ensure that he takes exactly one tablet from each of the bottle and at the same time he must avoid any wastage of the medicine.

Constraints:
1. Both bottles have equal number of tablets, say 30.
2. Tablets from both the bottles look exactly identical.
3. Medicine is very costly, so any kind of wastage is not affordable.

Problem Statement: How would you ensure that, in the above situation, you take exactly one tablet from each bottle, at the same time ensuring no wastage of the medicine.


Answer

- Take (1) Pill A from the bottle and add it to the 3 unknown pills. You now have (2) Pill A and (2) Pill B in your pile.
- Take each of the 4 pills and cut them in half.
- For each pill, put one of the halves in a pile on the right and one of the halves in a pile on the left.
- Each pile now contains 2 halves of Pill A and 2 halves of Pill B, which is the same as (1) Pill A and (1) Pill B in each pile.