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General discussionSolve the Maths Puzzle(114 posts)(114 posts)
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This is my favourite topic
Posted February 04, 2016
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?
Posted February 04, 2016
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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?
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?
Post edited February 04, 2016 by Hunter65536
Posted February 04, 2016
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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?
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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?
Correct. On top of or under the mother (depending of preference) would also be correct solutions. :-)

Solution to your riddle: The book has 103 pages.
Posted February 04, 2016
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zeogold: I could jump in to provide a question, if anybody needs me to. I can't guarantee it'll be all that hard, though.
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ZFR: Any time you're ready...
Your wish is my command, sahib.

So, gents, I happen to have a li'l cousin. In the year in which he was born, his brother was exactly one quarter of the age of their father, and now this brother is a third of the age of his uncle.
Oh, but I suppose it's only fair you know about my cousin as well, yes?
My cousin is a quarter as old as his dad, but never fear! In four more years, he'll be a quarter as old as his uncle will be!
So, how old is my cousin now?

Edit:
...somehow I was stupid enough not to refresh the page for an hour and entirely missed the posts that came before me.
..................well.
Post edited February 04, 2016 by zeogold
Posted February 05, 2016
Maybe I should've kept my mouth shut. I seem to have killed the thread!
Posted February 05, 2016
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zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!
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
Post edited February 05, 2016 by OneFiercePuppy
Posted February 05, 2016
low rated
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zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!
Now that's a brilliant suggestion,for the first sentence.
Post edited February 05, 2016 by Tauto
Posted February 06, 2016
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zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!
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Tauto: Now that's a brilliant suggestion,for the first sentence.
I could, but that probably wouldn't stop me from talking.
I mean, if you can talk without a brain, why wouldn't I be able to talk without a mouth?
Posted February 06, 2016
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zeogold: why wouldn't I be able to talk without a mouth?
I can't believe you typed that and didn't link to I Have No Mouth And I Must Scream.
Posted February 06, 2016
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zeogold: Maybe I should've kept my mouth shut. I seem to have killed the thread!
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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
Not quite, but you're close.
You mentioned that there are fractions involved. You didn't mention these fractions, I notice.
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zeogold: why wouldn't I be able to talk without a mouth?
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OneFiercePuppy: I can't believe you typed that and didn't link to I Have No Mouth And I Must Scream.
I just prefer to leave the aimless screaming to Tauto.
Post edited February 06, 2016 by zeogold
Posted February 15, 2016
This died again I see. So call this a bump in case anyone's still interested in the idea :)
Posted February 15, 2016
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.
Posted February 15, 2016
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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.
The inductive step doesn't work when n+1=2 since the two subsets you create have nothing in common.
Post edited February 15, 2016 by blarth
Posted February 15, 2016
Ick... Puzzles
Post edited February 15, 2016 by paladin181
Posted February 15, 2016
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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.
This was my thought as well. 0mA due to burning completely open from over current without a series resistor.
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General discussionSolve the Maths Puzzle(114 posts)(114 posts)
(114 posts)