yashsingh1/xdaforums huggingface dataset:
[Off-topic] stephj:
It's almost the same, but here the motive power is at the same end of both trains, not opposite ends as it was in the last puzzle, and here a few more steps are required.

I'll post the solution up in just over 24 hours, if nobody has got it by then.

O.K. Here it is:

--|
/
-----------------------------Ffffff-------------------Ppppppp-------

--|
/
--------------Ffffff----------------------------------Ppppppp-------

ff|
/
-Ffff------------------------------------------------Ppppppp-------

ff|
/
-Ffff----Ppppppp---------------------------------------------------

--|
/
-Ffff------------------------------------------------Pppppppff-----

ff|
/
-Ff--------------------------------------------------Pppppppff-----

ff|
/
-Ff-----Pppppppff--------------------------------------------------

--|
/
-Ff-----Pppppppffff------------------------------------------------

--|
/
-Ff-------------------------------------------Pppppppffff----------

Ff|
/
----------------------------------------------Pppppppffff----------

Ff|
/
--------Ppppppp--------------------------------------ffff----------

--|
/
--------Ppppppp------------------------------------Ffffff----------
[Off-topic] stephj:
A young boy had a bag of marbles stolen, which he reported to the police.

When questioned about the incident, the officer asked him, "Exactly how many marbles were in the bag?"

"I don't know." the boy replied, "If I counted them out in groups of two, three, four, five or six, there was always one left over, but if I counted them out in groups of seven, there was none left over."

What is the smallest number of marbles that could have been in the bag?
[Off-topic] 4silvertooth:
Is it 301?

#include <stdio.h>

/*
Counting in groups of 7 theres none left tells us that answer is multiple of 7
and counting in groups of 2,3,4,5,6 tells us that dividing the multiple of 7 theres
a remainder 1. So (7*i) (%) mod 2,3,4,5,6 will give us the answer.
*/

int main ()
{
for(int i=1;;i++)
{
if((7*i)%2==1 && (7*i)%3==1 && (7*i)%4==1 && (7*i)%5==1 && (7*i)%6==1)
{
printf("Minimum Marbles : %d\n",7*i);
break;
}
}
return 0;
}
[Off-topic] stephj:
The brute force approach works, and 301, is of course correct, but there is a smarter way to do it.

It's time to brush up on the Chinese Remainder Theorem. Break the numbers into their prime factors:-

2 = 2

3 = 3

4 = 2x2

5 = 5

6 = 2x3

The lowest common multiple of the above is 2x2x3x5 = 60

The smallest number having a remainder of 1, when divided by 2,3,4,5 and 6, apart from 1 itself, is therefore 60+1.

The answer we require is the smallest value of y such that (60y+1) is divisible by 7, which occurs when y reaches 5.
[Off-topic] huggs:
7

&10chars
[Off-topic] 4silvertooth:
But if we count in numer of 4 there will be 3 left.

For 5 - 2.
[Off-topic] stephj:
In the following multiplication cryptarithm, the asterisks represent the prime digits 2,3,5 or 7.

Surprisingly there is only one solution.

***
**
---
****
****
-----
*****
[Off-topic] stephj:
The above cryptarithm still runs, but meanwhile...........

A wealthy sultan decrees that after his death, his estate should be divided between his sons.

How many sons has he?
[Off-topic] Dirk:
Whilst the genius' are working on that..

... a nice easy Riddle:

Q) As a whole I am both safe and secure. Behead me and I become a place of meeting. Behead me again and I am the partner of ready. Restore me and I become the domain of beasts. What am I?

Edit: A completely unrelated, bonus trivia question: What is the most commonly played sport in the US?
[Off-topic] stephj:
Maybe a few pointers are required...................

In the puzzle in post #992 if A,B,C and D represent the digits of the solution, we have:

CCA x
BB
---
DBDA
DBDA
-----
DAACA

That tells you what 'A' is...............

The answer to the puzzle in post #993 relies on a rather pedantic use of english.
[Off-topic] stephj:
The penny has just dropped for Dirk's puzzle #994 above.

Considering I enjoy having a crack at the cryptic crossword in one of our British daily newspapers, I should have been here days ago.........

Answer: (S)(T)ABLE
[Off-topic] Dblfstr:
Stable

10 trolls

doh' ---^

---------- Post added at 11:15 AM ---------- Previous post was at 11:08 AM ----------

775
x 33
______
2325
23250
______
25575

a=5

b=3

c=7

d=2
[Off-topic] Dirk:
Stable is correct!

So what is the most commonly played sport in the US?
[Off-topic] Dblfstr:
Poke-her (poker)

Well, that is what I play the most....

But I would wager that baseball is the most common. It is our "favorite pastime"
[Off-topic] stephj:
Correct. Well Done.
[Off-topic] Dirk:
You'd think it's that obvious wouldn't you? But it's not! :eek:

Anyone else..?
[Off-topic] Dblfstr:
Did some research, it seems that this is a highly debated fact. Many, many results with varying degrees of certainty. I am not going to post my conclusions as I have already wagered a guess. But I think I have the top three "actual" most played sports in America. The number one was not even in the list of sports that popped into my mind at first.

Although, I think I had heard that before.
[Off-topic] Dirk:
He bought the 'numbers' for his house, '6', '0' and '0'.

Amirite?
[Off-topic] Archer:
Ahh.... clever
[Off-topic] deliriousDroid:
Your good. Respect

Sent from my SilverBullet 'HTCDesireZ running CM7'
[Off-topic] Dirk:
Q) How many months of the year have 30 days?

GO...!
[Off-topic] Archer:
:rolleyes:

All of them, except February.

This one's just as "hard" as that :p...

What's big enough to put a pig in, but small enough to hold in your hand?
[Off-topic] Tomdg07:
a blanket?

or maybe my stomach!!!

a bag?
[Off-topic] Dirk:
^^What he said i think, a blanket?

Here's a classic:

Q) What goes up the chimney down, but not down the chimney up?
[Off-topic] Archer:
An umbrella ;)

And no-one's got the earlier one yet.
[Off-topic] Tomdg07:
x2 on umbrella, but which one did we not get yet?
[Off-topic] Archer:
That ^^
[Off-topic] Tomdg07:
i got it, a pen!!!
[Off-topic] Archer:
Bingo! :)

But what get's wetter, the more it dries?
[Off-topic] waxworkz:
A towel? :)
[Off-topic] Archer:
Yesm :)

WE NEED MORE!!
[Off-topic] deliriousDroid:
The person who makes it has no need for it. The person who purchases it does not use it. The person who does use it does not know he or she is. What is it?

Sent from my SilverBullet 'HTCDesireZ running CM7'
[Off-topic] DJHollywood03:
A coffin? <------ Is that correct? Hummmmm Hope i dont see one anytime soon... Oh wait i will be dead so i wont see it!!
[Off-topic] DJHollywood03:
What is light as a feather but the strongest man in the world cannot hold for more than a few minutes?
[Off-topic] sygeek:
Brrrrrreath.

Okay, here's one:
[Off-topic] Dirk:
Rain?

Ok, here's tonight's puzzler:

A man goes out hunting for bear. Leaving his shelter in the morning he travels five miles due South. Not finding any bear by lunchtime, he changes direction and travels five miles due East. By mid afternoon he still has had no luck finding bear and decides to start heading home, turning due North.

Five miles later he is surprised to find himself back at his shelter!

Explain.....GO!
[Off-topic] xaccers:
He's surprised because he forgot his home was at the north pole.
[Off-topic] Dirk:
It's certainly not what you'd expect!
[Off-topic] Dblfstr:
I'm starting to enjoy the OT forum.
[Off-topic] stephj:
Time to terminate this one. A previous clue was, that the answer is given by a pedantic use of English.

The answer is two.

Any more and the word 'between' would be 'amongst'.

Meanwhile: It would appear Dirk's how old are you? thread, has been boarded up by the moderators.

So I'll put my reply in as a puzzle instead.

Next year, if you take my age in hexadecimal, it is the reverse of the two decimal digits.

How old am I this year?
[Off-topic] Dblfstr:
34...

Now this is my age:

Edit: To elaborate.

Using 35 as the hex input. Converting, the decimal equivalent to that number (35) is 53 - the reverse of 35. Since 35 is your age next year, this year you are 34.
[Off-topic] Archer:
I misread your question and got 52. I think the above is correct.
[Off-topic] Dblfstr:
My dyslexia almost got the best of me also (52). It was confusing the way it was worded, yet I do not think it was a bad wording of the problem.
[Off-topic] Dirk:
Anybody want to hazard a guess at most commonly played sport in the US?
[Off-topic] Archer:
Naked twister.
[Off-topic] MacaronyMax:
Eating? :rolleyes::p

But would have said American foot ball, Or even propper football?
[Off-topic] Dblfstr:
Tomorrow it WILL be eating AND football*.

*American Football
[Off-topic] Dirk:
Naked twister while eating?

Ok, i'll give it up. It's actually Ten-pin Bowling! :)
[Off-topic] Dblfstr:
Nobody wanted to attempt the guess my age?

Does the image I posted even show up?
[Off-topic] MacaronyMax:
Really? Sweet :p
[Off-topic] Dirk:
I see the image, i just have no idea what language it's in! :p

Minds immeasurably superior to mine will probably be along in 3..2..1..
[Off-topic] stephj:
Decimal 53 = Hex 35, but that is next year, so I'm 52.
[Off-topic] stephj:
A number with two identical digits is multiplied by 99.

1) Easy: - What is the four digit product if the third digit is a 5

Easier: - Use a calculator!

2)Harder: - Develop a method for providing the answer to generic questions of the form "What is the four digit product if the (1st,2nd,3rd, or 4th) digit is a (0 to 9)"

Then turn the calculator off. You don't need it.

The first or last digit can't be a zero and the second and third digit can't be a 9
[Off-topic] Dblfstr:
But you said "My age in Hex", so I took it as that number in hex was your age. Either way, conversion was right.

---------- Post added at 11:17 AM ---------- Previous post was at 11:10 AM ----------

are you looking for,

nn*99 = NOPQ, where

N = n,

O = n-1,

P = 9-n,

Q = 10-n

---------- Post added at 11:19 AM ---------- Previous post was at 11:17 AM ----------

Nobody?

10 char
[Off-topic] stephj:
Correct! With that remembered, you can regenerate the other numbers in the solution when given just one of the digits and its position.

Well Done!
[Off-topic] Dblfstr:
---------- Post added at 08:40 AM ---------- Previous post was at 08:34 AM ----------

Ok, since no one wanted to try:

The summation of 'i' from i = 1 to n where n = 8, quantity minus 6 =

[n(n+1) / 2] - 6 = [8(8+1) / 2] - 6 = [(8*9)/2] - 6 = (72/2) - 6 = 36 - 6 = 30
[Off-topic] stephj:
A grill can hold two slices of toast and it takes 30 seconds to toast one side.

\/\/\/\/\/\/\/\/\/\/\/\/\/

============ ============
--------------------------

How long does it take to toast three slices?

P.S. Be quick about it, I haven't got all day!
[Off-topic] Archer:
90 seconds :) I believe this has already been asked a while back.
[Off-topic] stephj:
So it was, next time I'd better search the thread properly!
[Off-topic] Archer:
103 pages of mixed up puzzles? That sounds like a puzzle in itself!

You should watch the off-topic thread. Some pictures have URLs on them so people visit the URL and browse around, find a different picture they like and then post that. Someone else comes along, sees the second picture and goes to the same URL to find a different picture. After a couple of days of doing that you end up with the same first picture all over again! It's like the x degrees of separation, but with lolcats and 9gag.
[Off-topic] stephj:
There is a five digit number, such that, if you follow it by a one, it is three times greater than if you precede it by a one.

i.e. ABCDE1 = 3 x 1ABCDE, where ABCDE is the five digit number in question.

One solution is to write a program to brute force the problem, but it is also possible to solve it using analysis alone.

So, what's the number?

Edit:- I'll drop a clue. Plain old algebra will more than suffice.

x * 10 + 1 = 3 * (100000 + x)

Come on! This is 8th grade stuff. (UK Year 9)...........
[Off-topic] stephj:
Puzzle #1026 is still open. You have had over a month, including the clue.

Meanwhile:-

The King's coinsmith is a crafty old so-and-so.

Each batch of coins are rolled up with each roll containing 100 coins.

He has already slipped one duff coin at random into each roll.

The King is a suspicious old so-an-so.

He opens each of 100 rolls of coins, selects a coin at random, and has it tested for purity.

What are the chances, as a percentage, that the coinsmith will be dragged off to the dungeons for an indefinite period?
[Off-topic] Dblfstr:
ABCDE1 = 3*1ABCDE

10x + 1 = 3*(100,000 + x)

10x + 1 = 300,000 + 3x

7x + 1 = 300,000

7x = 299,999

x = 299,999/7

x = 42857

428571 = 3*(142857)

ABCDE = 42857

---------- Post added at 02:33 PM ---------- Previous post was at 01:58 PM ----------

His chances of getting caught are 1 out of 100, if there were only one roll.

1/100 = 0.01 = 1%

But there are 100 rolls.

Bernoulli trial:

P(k) = p^k * q^(n-k)

Where p is the probability of failure (getting caught)

n is the number of trials

q is the probability of success, or not getting caught; 1-p

and k is number of failures (in this case 1 failure is enough)

p= 0.01

n=100

q = 0.99

k = 1

P(1) = 0.01^1 * 0.99^(100-1)

P(1) = 0.00369 = 0.37%

This may be wrong, I cannot remember all of my distributions and combination functions. This is under the assumption of exactly one fail out of 100 trys.
[Off-topic] stephj:
Good stuff on both counts!

The second one is almost right but it is near enough.

The chance of the coinsmith getting away with it is 37%. The chance of being dragged off to the dungeons is 63%.

Were the number of coins to increase and the number of samples to increase likewise, i.e 1000 coins per roll and one coin being sampled from each of all 1000 rolls etc. etc. 10,000 100,000 1,000,000 and so on, the result tends towards the reciprocal of e = 0.367879441

Well Done!!
[Off-topic] stephj:
A census taker asks the occupant of a house how many people live there and their ages.

The occupant says, “There are three of us. Our ages multiplied together equal 1296, and our ages added together equal the house number.”

The census taker notes the house number.

The occupant then adds, “You will also need to know that one of us is as old as the number of cars in the street. The census taker counts the cars, and then fills in the form with the ages of the occupants.

The ages of the occupants are whole numbers of years.

What’s the house number?
[Off-topic] stephj:
Post #1030 is still running. Any takers?

Meanwhile,

How many 2 inch matches does it take to pave a square yard in equal squares of 2 inch sides?
[Off-topic] Dblfstr:
Ages 6, 9, 24

House Number = 39

Seems like we needed to know at least the number of cars in the street.

The ages needed to be divisible by a factor of 3, since the final result was.

Break the number into segments 1296 to 12 9 6. three numbers decreasing by a factor of 3. 3(3*3) = 39. That is just something I noticed.
[Off-topic] stephj:
If the house number were 39, then the ages of the occupants would be 6, 9 and 24, but the census taker would not need to know the number of cars in the street.

Hint: House number 39 is a unique solution.

No they don't.
[Off-topic] Dblfstr:
Like a match to start a fire?
[Off-topic] Archer:
There isn't "an answer" to this. For example:

53 = 2 + 24 + 27

43 = 4 + 12 + 27

39 = 6 + 9 + 24

but they all multiply to 1296. Without knowing the number of cars in the street we're just guessing.

Edit: Actually, if the number of cars in the street was 27 then the above proves that there's still multiple answers. If you're looking for a specific answer then there must be only 1 way to calculate it when you know the number of cars.
[Off-topic] stephj:
Yes.

If the matches are two inches long scale this up to cover a paving block a yard across each side, with two inch squares.
[Off-topic] stephj:
There is only one house that has more than one solution that adds up to that house number.

The census taker needs the number of cars to decide which is the correct solution, we don't. We only need the house number with two solutions.

Don't forget that 1 is also a factor of 1296
[Off-topic] Dblfstr:
I see, per Archers answers. I didn't 'truly' figure this out mathematically. It was an arbitrary answer.

But I am missing something. 5 unknowns with three equations

age 1 = x

age 2 = y

age 3 = z

House number = H

Number of cars = C

x*y*z = 1296

x+y+z = H

x = C

Where we have, with what is given, no unique solution.

With 1 as a factor we have:

1 + 27 + 48 = 72

1 + 12 + 108 = 121

1 + 24 + 54 = 79

---------- Post added at 01:38 PM ---------- Previous post was at 12:56 PM ----------

One yard is 36 inches, broken into 2 inch (square) parts, there are 18 squares .

For a square yard, there are 18 squares horizontally, and vertically.

Further, each square has four sides, yet each square shares its sides with another square.

Considering each square to have at least one unique side, we can see that there are 18 total unique sides per horizontal and per vertical. plus one square that has two unique sides at the end. This leaves 19 total sides, or segments.

There are 19 segments horizontally per 18 squares.

There are 19 segments vertically per 18 squares.

(19*18) + (19*18) =

2*(19*18) = 684

Represented as such:

4 matches to start the top left corner (square):

3 matches to complete each additional square along the top row(17 squares)

3 matches to complete each additional square along the left side (17 squares)

The remaining 17*17 2" square sections will require 2 matches each. 17^2 * 2

4 + 2(3*17) + (12^2 *2) = 684

Visually:

4	3	3	3	3	3	3	3	3	3	3	3	3	3	3	3	3	3
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
3	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2	2
[Off-topic] stephj:
684 is correct. Well done!

The generic equation of:-

4+2*3*(x-1)+2*(x-1)*(x-1)

simplifies to:-

2x²+2x

Regarding the census taker problem, there are 42 possible combinations of ages, although some are much less likely than others:

1*1*1296 with a sum of 1298 for example.

40 of the 42 solutions have a unique sum of all three ages.

So, which sum appears twice? That is the house number.

If the house number was any different to what it is, then the census taker would know the ages of all three occupants immediately. He needs the number of cars on the street in order to decide which of the two sets of three ages is the correct one, as there is no common age in both sets.

We don't need the number of cars in the street, as we do not need to know which set of ages is correct. We only need the house number.
[Off-topic] Dblfstr:
Bump more puzzles.
[Off-topic] stephj:
It is finally time to put the puzzle in post #1030 to bed.

Here is the 'C' code that will churn out all the 42 possibilities of three ages that multiply to 1296, and the sum of those three ages.

#include <stdio.h>
main()
{
int i,j,k;
for(i=1;i<1297;i++)
for(j=i;j<1297;j++)
for(k=j;k<1297;k++)
if(i*j*k==1296)
printf("%5d%5d%5d%5d\n",i,j,k,i+j+k);
}

Take the output from this, and load it into Excel or similar, then sort the data on the final column.

Thus:-

9	12	12	33
9	9	16	34
8	9	18	35
6	12	18	36
6	9	24	39
4	18	18	40
6	8	27	41
4	12	27	43
3	18	24	45
3	16	27	46
6	6	36	48
4	9	36	49
3	12	36	51
2	24	27	53
2	18	36	56
3	9	48	60
4	6	54	64
3	8	54	65
2	12	54	68
1	36	36	73
1	27	48	76
1	24	54	79
3	6	72	81
2	9	72	83
4	4	81	89
1	18	72	91
2	8	81	91
1	16	81	98
3	4	108	115
2	6	108	116
1	12	108	121
3	3	144	150
1	9	144	154
2	4	162	168
1	8	162	171
2	3	216	221
1	6	216	223
2	2	324	328
1	4	324	329
1	3	432	436
1	2	648	651
1	1	1296	1298

A lot of these are extremely unlike to occur in reality, but that's not the point. Scan down the last column, looking for the only value that occurs twice......

91 - That's the house number.

Were the sum of the ages to be any other value, then the census taker would know all the three ages instantly. He needs to know the number of cars, in order to decide whether the occupants are 1, 18 and 72 or 2, 8 and 81. The number of cars in the street is one of these six values, but we couldn't care less.
[Off-topic] stephj:
In a nearby village there are three pubs:

The White Swan, The Black Bull, and The Grey Goose.

On Sundays all the pubs are closed.

The White Swan opens five days a week; The Black Bull opens four days a week, and The Grey Goose open three days a week.

The White Swan is never open four day running; The Black Bull is never open three days running, and The Grey Goose is never open two days running.

During a particular three day period, The White Swan is closed on the third day, The Black Bull is closed on the second day, and The Grey Goose is closed on the first day.

On Saturdays and Mondays, at least one pub is closed.

The opening pattern is the same each week.

On which day are all the pubs open?
[Off-topic] roofrider:
Censored !! (wrong answer)

Correct?
[Off-topic] stephj:
The answer is not Monday, Saturday or Sunday. [Statements 3 and 7].

To prevent random guessing, you will have to provide, as part of your answer, the full rota for the week for all three pubs!
[Off-topic] Dblfstr:
Edit: cannot read!
[Off-topic] stephj:
Look at the column for Sunday.
[Off-topic] Dblfstr:
Crap, missed that part. :(

Resubmit:

Tuesday:

iff Friday is the first day (Grey Goose closed)

Saturday is the second day (Black Bull closed)

Sunday is the third day (White Swan closed)
[Off-topic] stephj:
That's much better! Well done.
[Off-topic] stephj:
1. Easy

A father is watching his son in a sports day race.

Competitors numbers have been issued consecutively from 1 upwards. 1,2,3,4,5........ as one would expect.

The father notices that the sum of the runners' numbers that are less than his son's is the same as the sum of the runners' numbers that are greater than his son's.

If there are less than ten runners in the race what number has his son been allocated?

2. Not quite so easy.

Later that month the father is running in a much bigger race, but he notices exactly the same thing occurs with the number he has been allocated.

He has been allocated a three digit number. What is it?
[Off-topic] Dblfstr:
Players are: 1, 2, 3, 4, 5, 6, 7, 8

Boys number is 6

1+2+3+4+5 = 15

7+8 = 15

---------- Post added at 12:50 PM ---------- Previous post was at 12:20 PM ----------

There are 288 runners 1,2,3..,288

Dad's number is 204

1 to 203 = (1+n)*(n/2) = 20,706

Sum 205 through 288 = 20,706
[Off-topic] stephj:
Correct!

They are the only 1 and 3 digit solutions to

x*(x+1)=2y² where both x and y are integers.

This also makes 288/204 a pretty good approximation for the square root of 2.
[Off-topic] Dblfstr:
Seems it is just me an you in here.

Next puzzle I will wait a while and let someone else try to answer. :D
[Off-topic] 4silvertooth:
Hi stephj dirk and everyone.... bring it on.....

Can't find any running posts.
[Off-topic] 4silvertooth:
Cryptography....

5+3+2=151022

9+2+4=183652

8+6+3=482466

5+4+5=202541

then

9+5+5=_________?
[Off-topic] K.A.:
454585

1st number x 2nd number

1st number x 3rd number

sum of both products minus the second number..