You must really like riddles, eh Bushra :p. Here’s another one:

Murder Mystery

A man was found murdered one Sunday morning. His wife immediately called the police.
The police questioned the wife and staff and was given these alibis:

The Wife said she was in bed reading a book.
The Cook claimed she cooking breakfast.
The Gardener claimed he was planting seeds.
The Maid claimed she was getting the mail.
The Butler claimed he was polishing the silver.

The police instantly arrested the murderer. Who did it and how did they know?

It was either the wife because why in the heck would anyone be in bed reading after they wake up...

Or the maid, who claimed to be fetching the mail on a Sunday. Which begs the question, perhaps she didn't collect it on the Saturday...but then why of all times, when the guy was getting murdered would she suddenly go grab the mail?

I'm gonna go with the maid getting the mail on Sunday as well.

Yep! Mail on Sunday seems to be it, In a house with so many servants its not that possible that they managed to collectively forget picking up mail on Saturday.

Here's one:

Steve, a party magician, is carrying three pieces of gold each piece weighing one kilogram.

On the way to a session he comes to a 5 km bridge which has a sign posted saying the bridge could hold only a maximum of 80 kilograms.

Steve weighs 78 kilograms and the gold weighs three kilograms.

He reads the sign and still safely crossed the bridge with all the gold.

How did he manage this?

p.s: He does not throw it anywhere(Eg: Across the bridge)

He juggled as he was walking across the bridge

Originally Posted by huzaifa47

Yep! Mail on Sunday seems to be it, In a house with so many servants its not that possible that they managed to collectively forget picking up mail on Saturday.

Here's one:

Steve, a party magician, is carrying three pieces of gold each piece weighing one kilogram.

On the way to a session he comes to a 5 km bridge which has a sign posted saying the bridge could hold only a maximum of 80 kilograms.

Steve weighs 78 kilograms and the gold weighs three kilograms.

He reads the sign and still safely crossed the bridge with all the gold.

How did he manage this?

p.s: He does not throw it anywhere(Eg: Across the bridge)

Assuming he can't really do magic...

He left one piece of gold on one side, moved across with the remaining two, deposited them on the opposite side, and went back for the third piece.

OR

He removed his shoes.

OR

He "Accio Firebolt"ed his ...Firebolt...if he really could do magic

OR

He went to the bathroom before crossing

I'm gonna go with juggling too...in case it is the answer, I'll try to make one up, lol. Hopefully it's not too technical for anyone here. (:

Suppose I have a 'magic black box' in which I insert a numbered card, and a number displays on the screen. Suppose also that you have as many cards as necessary, with any possible numbers you can imagine. You can put in as many different cards as you want.

You know beforehand, that the box is actually doing a calculation. There is a mystery polynomial, f(X) = a0 + a1X + a2X^2 + ... + anX^n. (You don't know what n is. It could be 0, or 1,000,000 or more) So the number on the card you put into the box, is what is substituted for X.

For instance, if I put the numbered card with "0" on it, then the screen shows a0 (because f(0) = a0 + a1(0) + ... an(0)^n = a0 (whatever that happens to be)).

What you do know, is all of the coefficients, a0, a1, etc. are POSITIVE INTEGERS (this is crucial). That means they are in the set {1,2,...}, no decimals, and no negative numbers.

The goal is, you want to figure out {a0,a1,...,an} (and in doing so, figure out what the mystery polynomial is).

So the question is: What is the smallest number of 'number cards' I must insert into the black box, in order to figure out every coefficient (and to know for sure that I have done so)?

Uh oh I killed it. ): The answer is 2 (regardless of how large n). If you'd like, try to think of why that is the case, or a method of solving for the coefficients in 2 steps.

Someone else please pose a riddle haha.

For 30 picarats!

Mice can reproduce extremely quickly. Assume a mouse will give birth once a month, so that in 12 months one mouse will have 12 babies. Each mouse will mature and can reproduce in 2 months. You just bought a mouse that has just been born. How many will you have in 10 months?

Originally Posted by Mowicz

Uh oh I killed it. ): The answer is 2 (regardless of how large n). If you'd like, try to think of why that is the case, or a method of solving for the coefficients in 2 steps.

Someone else please pose a riddle haha.

I was gonna post my answer of "wth?!"

I understood precisely none of it lol

Originally Posted by Mowicz

Uh oh I killed it. ): The answer is 2 (regardless of how large n). If you'd like, try to think of why that is the case, or a method of solving for the coefficients in 2 steps.

Someone else please pose a riddle haha.

Can you post why that is the case? I did solve the question but I always have a problem figuring out why that is the case for maths stuff!

Originally Posted by jc0390

For 30 picarats!

Mice can reproduce extremely quickly. Assume a mouse will give birth once a month, so that in 12 months one mouse will have 12 babies. Each mouse will mature and can reproduce in 2 months. You just bought a mouse that has just been born. How many will you have in 10 months?

One... Dot dot dot ?

Huzaifa: I'll post it in a day or two, try to let it sink in and think what important "questions" (or numbers) you could ask the box (:

Originally Posted by Mowicz

One... Dot dot dot ?

Correct

No takers eh?

The solution is 2...and to prove this is the smallest number, I have to illustrate that 1 will not guarantee a solution, as well as solve the problem in 2 steps.

Step 1: 1 will not work:

Suppose my polynomial is the simplest it can be...f(x) = 100. (Or you can use just about any number)
Now suppose I input a card with the number A on it. Regardless of what it is, f(A) = 100.

So now I have one of two trains of thought:

1) Either f(x) := 100
2) Some crazy coincidence lead to the polynomial magically coming out to 100 when I inputted this number.

How do I know which one it is? I have to put in another card B, with B =\= A...

Now f(B) = 100 also...so we know f(x) := 100.

This isn't a solution to the problem, it just shows that even in this simple scenario, it's impossible to do in general with only one guess. It requires at least 2 to be absolutely certain.

Step 2: It can be done in EXACTLY 2 steps:

To show this is the case, I'll give an algorithm for determining all the coefficients.

Suppose f(x) = A + Bx + Cx^2 + Dx^3 + ...

First step) Put in the number 1. The box displays A + B + C + D + ..., since:
f(1) = A + B(1) + C(1)^2 + D(1)^3
= A + B + C + D + ...

Now the important thing to note is that A,B,C, D and so on, are all positive integers. That means, A,B,C,D and so on, are all less than or equal to A + B + C + D + ... = f(1).

Second step) Count the number of digits in A + B + C + D + ... = f(1). Let's say it has n digits. Every one of A, B, C, and D are less than f(1) so they each have either the same, or fewer digits.

Now, plug in the number 1 with n zeroes after it. You'll be pleasantly surprised, and be able to see every digit of the polynomial written out as one big long number! (With possible zeroes in the middle).

Example: This may not be immediately clear, so let's do an explicit example, and show you that this works:
Let's say f(x) = 62 + 10283x + 28430928x^2 + 3829832x^3 (typed at random).

A) First, we try plugging in the card with 1 on it:
f(1) = 62 + 10283(1) + 28430928(1)^2 + 3829832(1)^3
= 32271105

So after the first card, the box displays "32271105."

B) Next, we count the number of digits in f(1) = 32271105...there are 8.
So now we plug in 1 with 8 zeroes after it, 100000000. Now look at what the box will display:
f(100000000) = 62 + 10283(100000000) + 28430928(100000000)^2 + 3829832(100000000)^3
= 62 + 1028300000000 + 284309280000000000000 000 + 382983200000000000000 0000000000

(Ahh! Big numbers)

The box will display the number 382983228430928000102 8300000062.


Notice anything? Cut the above number after every 8 digits (starting from the right) and you will have your coefficients:

3829832|28430928|0001 0283|00000062

f( x ) = 00000062 + 00010283x + 28430928x^2 + 3829832x^3

Now just remove leading zeroes and you'll have the exact polynomial you wanted. (:

We need more riddles. I haven't seen Greg or Joseph on for a long time!