Geralt_of_Rivia: Since nobody posted a new question here is a quick and easy one to pass the time until the next difficult one:

A mother is 21 years older than her daughter and in 6 years the daughter will be 5 times younger than her mother. Where is the father?

Hunter65536: Let's say daughter is x years old and mother is x+21 years old. So after 6 years their ages will be x+6 years and x+27 years. Upon solving 5x+30=x+27 we get x=-0.75.
9 months before she is born so it means the mother and father are doing it right now. Also the answer to question, inside the mother.

Here's my question

If 201 digits were used to number the pages of a book, starting from 1, how many pages does the book have?

ZFR: Any time you're ready...

zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!

zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!

zeogold: why wouldn't I be able to talk without a mouth?

OneFiercePuppy: Sorry, was busy getting butchered in XCOM. Hm, yes, math puzzle. Yes. Ah, very simple. I see that your trick was "in the year" instead of "on the day" so we have different birthdays and year fractions. Here you go, then.

Eight years ago, your cousin was born. His brother turned eight later that year, when their dad was still 32. Your cousins brother is now 15, and their uncle is 45. Now this is when it gets a little dark. Their dad was driving home from work late one day shortly after your cousin's birth, and drove off the road and hit a pole. He was killed instantly, and this is why your eight year old cousin is a quarter as old as his dad. Their uncle turned to drink having lost his brother, and will succumb to liver cirrhosis in three years at 48. In four years, your cousin will be twelve, one fourth the age his uncle was when he wasted away and died in a hospital, gasping with yellow eyes as the jaundice poisoned him.

You're a terrible person, zeogold, to ask problems like this.

EDIT: typo

OneFiercePuppy: I can't believe you typed that and didn't link to I Have No Mouth And I Must Scream.

dtgreene: Here is my proof that all positive integers are equal. Specifically, I show that, given every subset of the integers, every integer in that subset is equal. Your task is to find a mistake in this proof, which uses the technique of mathematical induction.

First, the base cast: given a set of 1 integer, it is obvious that every integer in the set is equal, as there are no two unequal integers in the set (because there are no two integers in the set in the first place).

Now, suppose that, for every set of n integers, every integer in the set is equal. We now take a set of n+1 integers. The first through nth integers in the set are equal because they form a set of n integers. The same can be said of the second through (n+1)st integers in the set. Hence, the first element of the set must equal every member of the intersection of those two subsets, and the same can be said of the (n+1)st element. Therefore, every integer in the set is equal.

Now, your task, as I mentioned above, is to find a mistake in this proof.

b_smith_81: I am an electronic technician, so I have a more than few thoughts on this.

Okay, an LED is a type of diode, so it has polarity. Then there are two cases to consider:

Case 1: The LED is forward biased (installed forwards). With a limiting resistor in series an LED will typically drop its rated voltage, usually 2V for a LED. The remaining voltage is carried through the resistor. Without that resistor, all the voltage will try to pass though the LED. Assuming the LED does not immediately burn out, then it will try to carry a very large current, potentially as much as the battery can produce (the current a diode carries grows exponentially as the voltage goes over its rating).

Case 2: The LED is reverse biased (installed backwards). There will be a few microamps of current due to reverse bias leakage (treated as 0 amps in practice). This is true unless the reverse breakdown voltage is exceeded; then the voltage overwhelms the p-n junction and potentially large amounts of current will flow, similar to the overvoltage scenario in the forward biased case.

The practical result: either the diode is carrying effectively no current due to either being reverse biased or burnt out.