• New TOR Mirror: suicidffbey666ur5gspccbcw2zc7yoat34wbybqa3boei6bysflbvqd.onion

  • Hey Guest,

    If you want to donate, we have a thread with updated donation options here at this link: About Donations

J

Julgran

Enlightened
Dec 15, 2021
1,428
Nice initiative! Here's one in video format:

 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
A census taker approaches a woman leaning on her gate and asks about her children.
She says, "I have three children and the product of their ages is 72."
He says, "How am I supposed to figure out their ages from that?"
She replies, "The sum of their ages is the number on this gate."
The census taker does some calculation, but then exclaims, "I'm sorry, but that's not enough information."
The woman had enough and says, "I don't have time for this; I have to see to my eldest child who is in bed with measles."
She slams the door and the census taker departs, satisfied.
How old are the children?

The solution is here, but I urge you to try it yourself first; it's not as difficult as it seems at first glance.

 
lostundead

lostundead

Student
Mar 18, 2021
192
A census taker approaches a woman leaning on her gate and asks about her children.
She says, "I have three children and the product of their ages is 72."
He says, "How am I supposed to figure out their ages from that?"
She replies, "The sum of their ages is the number on this gate."
The census taker does some calculation, but then exclaims, "I'm sorry, but that's not enough information."
The woman had enough and says, "I don't have time for this; I have to see to my eldest child who is in bed with measles."
She slams the door and the census taker departs, satisfied.
How old are the children?

The solution is here, but I urge you to try it yourself first; it's not as difficult as it seems at first glance.

That was fun. The measles bit confused me so much, I can't believe I fell for it. Thats like the oldest trick in the book for puzzles; throw in some unnessary specific information that beares no relevance whatsoever for the solution but leads people down the wrong path.
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
That was fun.
I remember once reading a more difficult version of this riddle, but I can't find it anymore.

Let n be a natural number. Prove that there are at least three primes between 10^n and 10^(n+1).
You may use Bertrand's postulate, which states that for every natural number m>1, there exists a prime p such that m<p<2m.
 
J

Julgran

Enlightened
Dec 15, 2021
1,428
I remember once reading a more difficult version of this riddle, but I can't find it anymore.

Let n be a natural number. Prove that there are at least three primes between 10^n and 10^(n+1).
You may use Bertrand's postulate, which states that for every natural number m>1, there exists a prime p such that m<p<2m.

I didn't know that headache would be my reason to end my life :smiling:

Anyhow, thank you for your input.
 
  • Like
Reactions: lostundead
lostundead

lostundead

Student
Mar 18, 2021
192
Let n be a natural number. Prove that there are at least three primes between 10^n and 10^(n+1).
You may use Bertrand's postulate, which states that for every natural number m>1, there exists a prime p such that m<p<2m.
Yeah, that's where the fun ends.
 
J

Julgran

Enlightened
Dec 15, 2021
1,428
Sorry. I can still delete it if it doesn't fit the thread, although it's really not as difficult as it looks.

No, no, no - it was just a joke about my own inability to comprehend your mathematical challenge :wink:
 
stygal

stygal

low-wage worker
Oct 29, 2020
1,732
A census taker approaches a woman leaning on her gate and asks about her children.
She says, "I have three children and the product of their ages is 72."
He says, "How am I supposed to figure out their ages from that?"
She replies, "The sum of their ages is the number on this gate."
The census taker does some calculation, but then exclaims, "I'm sorry, but that's not enough information."
The woman had enough and says, "I don't have time for this; I have to see to my eldest child who is in bed with measles."
She slams the door and the census taker departs, satisfied.
How old are the children?
I came up with different solutions - some of which mentioned in the Wiki: 2,4,9 and 3,4,6 as I thought there had to be an eldest child which still had to be in the age range to get measles...also where is the number on the gate even mentioned ? Confused me a bit.
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
I came up with different solutions - some of which mentioned in the Wiki: 2,4,9 and 3,4,6 as I thought there had to be an eldest child which still had to be in the age range to get measles...also where is the number on the gate even mentioned ? Confused me a bit.
Measles can affect people of any age. The brilliancy of the riddle lies exactly in the fact that you don't need to know the number on the gate, just that knowing the number is not sufficient. "Because the census taker knew the total (from the number on the gate) but said that he had insufficient information to give a definitive answer, there must be more than one solution with the same total." This is not the case with the answers you propose.
 
  • Like
Reactions: little helpers
stygal

stygal

low-wage worker
Oct 29, 2020
1,732
Measles can affect people of any age. The brilliancy of the riddle lies exactly in the fact that you don't need to know the number on the gate, just that knowing the number is not sufficient. "Because the census taker knew the total (from the number on the gate) but said that he had insufficient information to give a definitive answer, there must be more than one solution with the same total." This is not the case with the answers you propose.
Guess I got caught up in the words "eldest" child as if there was only one. Plus where I'm from measles are considered a children's disease which of course can affect all ages as well - I just thought that would be valuable information ^^ But I re-read the Wiki and understand now that what the census said was important and that he had already solved it only wasn't sure which one was the actual one
 
Last edited:
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Guess I got caught up in the words "eldest" child as if there was only one. Plus where I'm from measles are considered a children's disease which of course can affect all ages as well - I just thought that would be valuable information ^^ So I'm still not sure why the Wiki says two of the solutions are the "right ones" when in fact then all given solutions could be true?
The first line of the riddle tells us that the woman is leaning on her gate when the census taker approaches her. This means that he is able to see the number on the gate. Let's say the number on the gate was 3+4+6=13. In that case, he would have left right there and then, because there is only one set of ages that adds up to this number. Since he said that he has not enough information, it means that there must be more than one set of ages with the same total. In this case, this is only fulfilled by 2+6+6=14=3+3+8. When the woman now tells him that she has an eldest child, the census taker can rule out the triple (2,6,6) and is therefore only left with (3,3,8).
 
Last edited:
  • Like
Reactions: lostundead
stygal

stygal

low-wage worker
Oct 29, 2020
1,732
The first line of the riddle tells us that the woman is leaning on her gate when the census taker approaches her. This means that he is able to see the number on the gate. Let's say the number on the gate was 3+4+6=13. In that case, he would have left right there and then, because there is only one set of ages that adds up to this number. Since he said that he has not enough information, it means that there must be more than one set of ages with the same total. In this case, this is only fulfilled by 2+6+6=14=3+3+8. When the woman now tells him that she has an eldest child, the census maker can rule out the triple (2,6,6) and is therefore only left with (3,3,8).
Yes, I get it now - I just didn't know he had already solved the riddle and only needed to confirm which one of the two answers was the right one. I thought he didn't solve it and asked for more clues out of desperation - lol.
The 3.3.8 is definitely the one.
Thanks, for clearing it up and sticking with me - rather slow with this.
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
rather slow with this.
This is not a competition. The most important thing is that the riddle provided you with an opportunity for some pleasant divertissement.
 
  • Like
Reactions: stygal
lostundead

lostundead

Student
Mar 18, 2021
192
Let n be a natural number. Prove that there are at least three primes between 10^n and 10^(n+1).
You may use Bertrand's postulate, which states that for every natural number m>1, there exists a prime p such that m<p<2m.
Do you need an integral to solve this one?
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Yeah I see now that it doesn't make proper sense. Could you solve it please? I've had enough math for this year.
You've essentially got the solution in the first line you wrote.
We apply Bertrand's postulate three times to obtain three primes p', p'', p''' which fulfill the following inequalities:
i) 10^n<p'<2*10^n,
ii) 2*10^n<p''<4*10^n,
iii) 4*10^n<p'''<8*10^n.
By combining them we arrive at the following:
10^n<p'< 2*10^n<p''<4*10^n<p'''<8*10^n<10*10^n=10^(n+1).
In other words, we found three primes between 10^n and 10^(n+1).
 
lostundead

lostundead

Student
Mar 18, 2021
192
You've essentially got the solution in the first line you wrote.
We apply Bertrand's postulate three times to obtain three primes p', p'', p''' which fulfill the following inequalities:
i) 10^n<p'<2*10^n,
ii) 2*10^n<p''<4*10^n,
iii) 4*10^n<p'''<8*10^n.
By combining them we arrive at the following:
10^n<p'< 2*10^n<p''<4*10^n<p'''<8*10^n<10*10^n=10^(n+1).
In other words, we found three primes between 10^n and 10^(n+1).
Oh, gotcha. The second part of my calculation wasn't necessary, but it's correct nonetheless because I proved that 8*(10^n) is smaller than or equal to 10^(n+1).
 
GenesAndEnvironment

GenesAndEnvironment

Autistic loser
Jan 26, 2021
5,743
A powerful potion, to heal all ills. What is this substance, that often kills?
Not black nor green, on the wiki it is seen.
Not sweet nor sour, and will shorten the hour.

He types with great haste, "That fella is based!".
He's curved but straight, "Why don't I have a mate?".
Thousands of posts, neurotypicals he roasts.
"Are you a woman? Let's meet!", who is this vile NEET?
 
T

tardis

Member
Sep 7, 2019
73
I made this thread out of boredom, I wanted us to share riddles and brain games and try to find an answer, starting with me, does anyone know the answer of pic related? hahaha
View attachment 82081
15?

A census taker approaches a woman leaning on her gate and asks about her children.
She says, "I have three children and the product of their ages is 72."
He says, "How am I supposed to figure out their ages from that?"
She replies, "The sum of their ages is the number on this gate."
The census taker does some calculation, but then exclaims, "I'm sorry, but that's not enough information."
The woman had enough and says, "I don't have time for this; I have to see to my eldest child who is in bed with measles."
She slams the door and the census taker departs, satisfied.
How old are the children?

The solution is here, but I urge you to try it yourself first; it's not as difficult as it seems at first glance.

my solution is to use 72 = 2^3 3^2 and note that 14 = 3+3+8 = 6+6+2 which are two solutions to the first two statements. Then when she says she has a (unique) oldest it must be the 8 year old. So 3,3,8. I'm too lazy to check 14 is the only ambiguous sum of possible ages.

Edit: I checked the solution and this was right, but I never verified 14 is the only ambiguous solution from the first two statements.

I remember once reading a more difficult version of this riddle, but I can't find it anymore.

Let n be a natural number. Prove that there are at least three primes between 10^n and 10^(n+1).
You may use Bertrand's postulate, which states that for every natural number m>1, there exists a prime p such that m<p<2m.

you just partition numbers between 10^n and 10^(n+1) into
the intervals:

[10^n, 2*10^n], [2*10^n+1, 4*10^n+2], and then [4*10^n+3,10^(n+1)]

then you just have to show 10^(n+1) > 8*10^n+6 which holds when n>0. and you have
your proof from Bertrand's postulate.

Edit: Just realized you don't need to be adding ones to the intervals. So this will hold whenever b^(n+1) >= 8*b^n. In particular it holds for all n>0 whenever b>=8.
 
Last edited:

Similar threads

Ayyoooooo._
Replies
11
Views
483
Suicide Discussion
rozeske
R
Silent_cries
Replies
31
Views
417
Forum Games
whimsyaspie
whimsyaspie
DesperateOne
Replies
39
Views
706
Suicide Discussion
momento.mori
M