yashsingh1/xdaforums huggingface dataset:
[Off-topic] stamatis:
ok i see,so it would be 3+6+3

but i am not complete sure that the corner balls and the others will not touch the center one......hmmm....let me think about this a little further......

ok with a shape i see that in order for corner ball to touch the center one the distance between those two centers must be 2r(considering "r" the radious[<- not sure for the word] of the ball, but with the pythagorean theorem that distance is 8^(1/2)r

and 8^(1/2)r > 2r so neither the corner nor the between balls touch....and you are right!:):)

i admit that my thought was ruch and incomplete....

thanks
[Off-topic] Dblfstr:
To what is this referring?
[Off-topic] azzledazzle:
After the maths you have to be left with 6 not any other number
[Off-topic] Dblfstr:
x= (6-9)

y = (9-10)

z = (40-50)

x+y-z=

(6-9)+(9-10)-(40-50)

-3-1+10=6

:D

Don't know if that is what you are looking for. But it satisfies your requirement.
[Off-topic] azzledazzle:
Ive no idea what you just did lol, Maths was never my forte, I just googled it :)

The solution is:

SIX – 9 (IX) = S

9 (IX) – 10 (X) = I

40 (XL) – 50 (L) = X

===>SIX<===
[Off-topic] prototype7:
lol
[Off-topic] stephj:
A man goes into a shop to buy four items and decides to keep a running total of the value by adding them together on his pocket calculator.

Unfortunately his numeracy skills leave a little to be desired, because instead of pressing the plus (+) button on the calculator he inadvertantly presses the multiply (x) button instead.

When he gets to the cash desk the calculator displays 7.11

The operator scans the four items and the till displays the total for the four items as 7.11 (Pounds, Dollars Euros - Take your pick)

He pays for the items and leaves the store oblivious to his error.

What was the cost of each of the four items?
[Off-topic] Dblfstr:
Are you saying that the sum and product of the four numbers are 7.11?

Also, can we round to the nearest hundredth (for the product)?

If so, here is a solution:

2.22 + 2.2 + 1.94 + .75 = 7.11

2.22 + 2.2 + 1.94 + .75 = 7.106 ~ 7.11

2.21 + 2.2 + 1.95 + .75 = 7.11

2.21 + 2.2 + 1.95 + .75 = 7.1106 ~ 7.11
[Off-topic] stephj:
Yes.

No, the product is 7.11 exactly. Not a fraction of a penny more or less.
[Off-topic] stephj:
Maybe it's time to drop an analytical pointer or two.

The decimal points tend to throw you off the scent somewhat. Because the product is a multiple of four amounts of pounds/pence dollars/cents euros/cents whatever, each with two decimal places, then the resultant product must have eight decimal places.

7.11 has only two, so what happened? .........The last six digits of the product are zero!

Convert the amounts to pence/cents and we have four three digit numbers that multiply to 711,000,000

The prime factors of 711000000 are 2.2.2.2.2.2.3.3.5.5.5.5.5.5.79

One of the amounts is a multiple of 79 pence/cents all the numbers must use the all the above factors between them, otherwise the product can never equal 711000000.
[Off-topic] stephj:
Time to put this one to bed. A brute force attack will solve it in a few seconds.

#include <stdio.h>

main()
{
unsigned long i,j,k,l;

for(i=1;i<=355;i++)
for(j=i;j<=355;j++)
for(k=j;k<=355;k++)
for(l=k;l<=355;l++)
if(i*j*k*l==711000000 && i+j+k+l==711)
{
printf("%lu,%lu,%lu,%lu\n",i,j,k,l);
return(0);
}
}

Output:
120,125,150,316

The amounts are (in Sterling), £1.20, £1.25, £1.50 and £3.16

You have two hourglasses that run for four and seven minutes exactly.

How do you measure exactly nine minutes?
[Off-topic] Dblfstr:
Looks so simple when you do it that way.

Measure 9 minutes.

1. Start both hour glasses at the same time.

2. After 4 minutes (end of 4* glass), restart the 4* glass. - Total 4 Minutes

3. After 7 minutes (end of 7* glass) flip 7* glass over. - Total 7 minutes.

4. After 4* glass has ended, re-flip 7* glass. - Total 8 minutes

5. 7* glass will run for one more minute. - Total 9 minutes.

At step four the 7 glass will have been running for one minute (totaling 8). Once flipped it will run for one more minute exactly (totaling 9).
[Off-topic] Dirk:
I'm going to rename this thread the ** Off-Topic Give Yourself A Headache Thread **

Everyone here makes me feel like a dumbass! :(
[Off-topic] stephj:
In the words of a great British seaside children's entertainer, "That's the way to do it"

Well done.
[Off-topic] Dirk:
FTFY!

:p

Haven't tried a lateral thinking puzzle on you guys for a while. I'll have to see if i can remember any that i haven't used yet.

Edit: Okay, here's an easy one to get us started:

A man was pushing his car. He stopped when he reached a hotel, at which point he knew he was about to go bankrupt. Why?

And a slightly harder one:

An Arab sheikh tells his two sons that they are to race their camels to a distant city to see who will inherit his fortune. The son whose camel arrives last will win. After riding aimlessly for days, the brothers stop to ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to their destination.

Why?
[Off-topic] Dblfstr:
The man was pushing his car because the car had no power of its own. The hotel at which he stopped was owned by someone else. Clearly he hadn't enough money to pay the rent or barely enough, so he would be bankrupt. This happens sometimes in Monopoly.

I will have to think about the other one...
[Off-topic] Dblfstr:
I must say that this one was hard. I actually wasted some work time thinking about it.

I initially thought that the sheikh was perhaps in debt, or in war, something negative that neither son would want to inherit. Yet, it seems that they were trying to get there last, meaning they wanted to inherit it. So initially I was wrong.

The wise man told them something that spurred them to race as fast as they can; again I thought this was due to debt, etc. However, this was still too obvious, and clearly the sons would have been aware of that fact.

They two sons wanted to inherit the fortune. Neither would dare be the first to arrive, so they wandered aimlessly hoping the other would give in. This of course would never happen, not until one died...

The wise man, being very wise, must have given them a solution. It was stated that "The son whose camel arrives last will win." not the son who arrives last. Alas, if the two sons switched camels, they would indeed need to race to the finish line to have the other brother's camel arrive first, and their camel arrive last - making them the winner.

The wise man instructed them to switch camels.
[Off-topic] stephj:
You have two ropes.

They are not the same length or width as each other, in fact they are not uniform themselves, as the width varies along the rope.

The only property they both have, is that they will burn from one end to the other in exactly one hour.

Use them to measure 45 minutes.
[Off-topic] Dblfstr:
Dang, I have 30 minutes so far...

Edit:

Can I alter the rope?

Edit 2: Never mind I got it.

At first I was like:

Slit one rope down the middle (lengthwise) to make two identical ropes.

Light the unchanged rope at both ends (should burn for 30 minutes)

When that rope is burnt, light one half of the altered rope at both ends (should burn for 15 minutes)

Use the other half of the altered rope to make knots to kill time for 45 minutes. :D

But then I was like:

To measure 45 minutes.

Light both ends of one rope and one end of the other rope.

After the rope lit at both ends has burned completely, 30 minutes have elapsed.

At this point, the other rope has also been burning for 30 minutes;

light the other end of that rope and it will continue to burn for an additional 15 minutes.
[Off-topic] stephj:
And the second one is correct!
[Off-topic] Dblfstr:
Dee Septor, the famous magician, claimed to be able to throw a ping-pong ball so that it would go a short distance, come to a complete stop, and then reverse itself. He also added that he would not bounce the ball off any object, or tie anything to it. How could he perform this trick?
[Off-topic] stephj:
Back to a spot of number bashing.

There are several perfect squares that are also palendromic numbers.

11x11 = 121

22x22 = 484

26x26 = 676

101x101=10201

So what is the smallest perfect square with an even number of digits?
[Off-topic] jaszek:
4x4=16

Idk. 16 has 2 digits, which makes it an even number of digits.
[Off-topic] Dblfstr:
Yes, this is confusing. Are you looking for the smallest square with an even number of digits that is also a palindrome? Or just the smallest square with even digits.

4X4 is 16 which is even... but not a palindrome.
[Off-topic] stamatis:
What?

Are you really expecting an answer?I have heard about it before,the number you are asking for is 20 digits long.

You are evil:D

But in case you talked about palendromics numbers just to mislead then its 16=4x4,but I have sth in mind,give me some minutes.

Edit:I thought that it might be some number like this:0.yyy x 0.yyy but It seems I cant find even digits.
[Off-topic] prototype7:
I didn't know numbers could be paramedics :eek::p
[Off-topic] stamatis:
Oops I don't know why I wrote it,I mean palendromics:D
[Off-topic] erickxd:
sorry, i understand.

thanks :D
[Off-topic] stephj:
Sorry didn't quite make that clear enough.

So what is the smallest palendromic perfect square with an even number of digits?

@Stamatis16: If you are still looking for it at 20 digits you missed it miles ago.

Hint: There are no two digit answers, all palendromic two digit numbers are multiples of 11, itself prime, so the first palendromic square is 11x11 = 121 = 3 digits. No good.

OK. four digits:

Four digit squares are the squares of numbers 33 through to 99. < 33 result in three digits.

Ignore 40,50,60,70,80, and 90. The square ends in double zero and no square starts in double zero.

Also you can ignore all squares that don't start in a 1,2,5,6 or 9, as a perfect square has to end in one of these numbers. (Zero also applies but we've just knocked those out in the step above.)

To test all four digit perfect squares we only need to test the squares of the 35 numbers:-

33 to 39, 41 to 44, 64 to 69, 71 to 79, 81 to 83, and 95 to 99

It is none of these numbers, so try six digits using this or other methods...................
[Off-topic] Dblfstr:
Calculating now.

This is in excel btw, I do not have and other way to program at work.

Sub Macro4()
'
'Declare my i as long
Dim i As Long

'Set i to 100000
i = 100000

'set cell A1 = i
Range("a1") = i

'set cell B1 = i^2
Range("b1").Formula = "=A1*A1"

'Set cell D1 to check if the beginning is the same as the end, then in the number is even .
Range("D1").Formula = "=IF(AND(LEFT(B1,1)=RIGHT(B1,1),LEFT(B1,2)=RIGHT(B1,2),LEFT(B1,6)=RIGHT(B1,6)),IF(MOD(LEN(B1),2)=0,TRUE,FALSE))"

'Loop until my conditions are met, increment by 1
While Range("d1") = False
i = i + 1
Range("a1") = i

Wend

End Sub
[Off-topic] Archer:
There's this...

http://www.coderun.com/ide/​

But I'd recommend this...

https://compilr.com​

Much more project types on offer and it just seems a more professional all-round solution.
[Off-topic] stephj:
You can actually use Excel to solve the problem without using Macros:

I have given away the fact that we are looking for a six digit square.

Start Value

Cell A1: 317

Square it.

Cell B1: =A1*A1

Does the first digit of the square match the last?

Cell C1: =INT(B1/100000)=MOD(B1,10)

Does the second digit of the square match the next to the last?

Cell D1: =MOD(INT(B1/10000),10)=MOD(INT(B1/10),10)

Do the the two middle digits match each other?

Cell E1: =MOD(INT(B1/1000),10)=MOD(INT(B1/100),10)

Combination of all the three boolean values.

Cell G1: =AND(C1,D1,E1)

Make Cell B1 = 318

Cell B1: =A1+1

Copy B1 to G1 down to B2 to G2

Copy A2 to G2 all the way down to A683 to G683 (Numbers 317 to 999)

Find the TRUE value in Column G.
[Off-topic] Dirk:
Is the answer 42?
[Off-topic] Dblfstr:
Yep, but you gave it away. I also had an error in my code. Not to mention that I thought i was looking for a number with six digits to square. Not the square was 6 digits.

Anyway.

836*836 = 698896

Edit:

So I fixed my code. Did not realize I needed INT in there the check '='. Also I was starting at 100000, based on my previous comment. I now have the code fully working and it only takes a couple of seconds. This uses, though, the knowledge that the answer is 6 digits long. Any other answers would fail.

Sub Macro4()
'
On Error Resume Next

Dim i As Long
i = 3
Range("A1") = i
Range("B1").Formula = "=A1*A1"
Range("D1").Formula = "=IF(INT(LEFT(B1,3))=INT((MID(B1,4,(LEN(B1)-5)))+(MID(B1,5,(LEN(B1)-5))*10)+(RIGHT(B1,1)*100)),TRUE,FALSE)"

While Range("d1") = False
i = i + 1
Range("a1") = i
Wend

End Sub
[Off-topic] stephj:
Correct! Well done.

Another way of doing it is to run this:

#include <stdio.h>
#include <string.h>

main()
{
unsigned long i,j,k,palendromic;

char number[20];

for(i=4;i<65535;i++)
{
sprintf(number,"%lu",i*i);
j=strlen(number);
palendromic=1;

for(k=0;k<(j/2);k++)
{
if(number[k] != number[j-k-1])
palendromic=0;
}
if(palendromic)
printf("%lu,%lu\n",i,i*i);
}
}

The 'k' loop manages to deal with any length square. If the number of digits are odd, it ignores the middle one.

The output is :

11,121
22,484
26,676
101,10201
111,12321
121,14641
202,40804
212,44944
264,69696
307,94249
836,698896
1001,1002001
1111,1234321
2002,4008004
2285,5221225
2636,6948496
10001,100020001
10101,102030201
10201,104060401
11011,121242121
11111,123454321
11211,125686521
20002,400080004
20102,404090404
22865,522808225
24846,617323716
30693,942060249

We have to stop at 65535 or the squared value will overflow an unsigned long.

In the entire list so far, there is only one square with an even number of digits.
[Off-topic] stephj:
Sorry Dirk. It might be the answer to life, the universe and everything, but not this time.

Life..... Don't talk to me about life............
[Off-topic] Dblfstr:
Yep, got my macro to figure it all out too.

Sub Macro4()
'
On Error Resume Next

Dim i As Long
i = 1
ActiveCell = i
Range("B1").Formula = "=A1*A1"
Range("D1").Formula = "=IF(LEN(B1)>=3,IF(INT(LEFT(B1,3))=INT((MID(B1,LEN(B1),1)*100)+(MID(B1,LEN(B1)-1,1)*10)+(MID(B1,LEN(B1)-2,1)*1)),TRUE,FALSE),IF(LEN(B1)<3,IF(LEFT(B1)=RIGHT(B1),TRUE,FALSE)))"

For i = 1 To 10000000

If Range("D1") = True Then

While Range("D1") = True
i = i + 1
ActiveCell.Resize(1, 3).Select
Selection.Copy
Selection.Insert Shift:=xlDown
Application.CutCopyMode = False
ActiveCell.Offset(1, 0).Select
ActiveCell.Select
ActiveCell = i + 1
Wend

While Range("d1") = False
i = i + 1
ActiveCell = i
Wend

End If

Next

End Sub

OUTPUT:

1	1
3	9
11	121
22	484
26	676
101	10201
111	12321
121	14641
202	40804
212	44944
264	69696
307	94249
836	698896
1001	1002001

---------- Post added at 02:38 PM ---------- Previous post was at 02:05 PM ----------

anyone
[Off-topic] ssojyeti2:
Maybe attach a string to it lol?
[Off-topic] jonah1234:
He can't attach the ball to anything

Sent from my AOKP HTC Thunderbolt from Tapatalk 2.4
[Off-topic] cor4twenty:
Throw it straight up in the air : D
[Off-topic] Dblfstr:
Indeed. Good Job.
[Off-topic] ssojyeti2:
Damn. Should have got that one:p
[Off-topic] Dirk:
A quick, easy one one:

Q) You're in a race. You overtake the person in 2nd place. What position are you in?
[Off-topic] Dirk:
Correct!

Not so much a puzzle as an observation... what have i done to all my ** threads ** this afternoon?
[Off-topic] Dblfstr:
Changed the name (took away the statement in parentheses).

/stalker
[Off-topic] Dirk:
That too, lol...but i was so bored @ work i thematically tarted up the OP's. The one place nobody ever looks! :rolleyes:
[Off-topic] Dirk:
A new puzzle..

Q) Five XDA mods race to Off-Topic to close a thread. Mr Clown gets here before M_T_M but behind CTM. Archer gets here before TeeJay but behind M_T_M.

In what order do they get here?
[Off-topic] Dirk:
Well played!
[Off-topic] stephj:
Mr Smith always comes home on the same train, which arrives at the station at 5:00pm.

His wife picks him up in the car. She sets off from home, and drives at a constant speed to arrive at the station at 5:00pm prompt, as the train is just arriving.

She picks him up and they drive home at the same constant speed.

One day, Mr Smith arrives on the earlier train at 4:00pm. He hasn't rung his wife to tell her, and as it is a nice day, he sets off walking toward home at a constant rate.

Mrs Smith, meanwhile, leaves home in the car at the usual time to get the station for 5.00pm.

They meet somewhere between home and the station, where she picks him up and they return home.

They arrive home 10 minutes earlier than usual.

How many minutes was Mr Smith walking before his wife picked him up?
[Off-topic] Dblfstr:
I have not worked this mathematically, well I started and got hung up somewhere.

So I decided to use logic instead!

As I said, I have not proved this mathematically but I think I have it.

If they arrived home 10 minutes earlier than normal, that is 10 minutes off the round trip. Which is 5 minutes off a one way trip. If the wife normally picks the husband up at 5:00, she must have picked him up five minutes earlier than normal, or 4:55 (one way trip minus 5 minutes).

If he started walking at 4:00, and was picked up at 4:55, then he walked for 55 minutes.

So my answer is 55 minutes.
[Off-topic] stephj:
55 minutes is correct. Well done.

The speed of walking, driving, and the distance from the railway station to home, are not mentioned, but are also not required to work out the correct answer.
[Off-topic] stephj:
You are in the middle of dealing out playing cards for a game of bridge, when the phone rings.

You place the partly dealt pack on the table, and leave to answer the phone in the next room.

When you return, neither you nor any of the other players can remember to whom the last card was dealt.

Without you or the other players counting any cards, how do you complete the deal, giving all four players exactly the same cards they would have had, had the deal not been interrupted.
[Off-topic] Dblfstr:
1. Have each player set their cards to the side.

2. Start dealing the rest of the cards as if you had just started dealing.

3. Wherever the last card falls, shift the new decks to the left until you have the deck that got the last card.

4. Merge the new decks with the ones you set aside.
[Off-topic] stephj:
That would work.

But there is a much easier way to do it.....................
[Off-topic] Dblfstr:
Deal backwards from the bottom starting with yourself.
[Off-topic] stephj:
And that's it. Well done!
[Off-topic] stephj:
On the City and Seaside Railway, the locomotive driver, has one final job to complete before his shift is over.

His task is to swap the positions of the two wagons, (US - freight cars), A and B, and return the locomotive to its starting position in the spur at the top left.

The locomotive can fit through the tunnel to the east, but the wagons can't. No flying switch manoeuvres are allowed, i.e. where wagons are unhitched from the locomotive as they are moving, and a member of the crew leaps off the train to throw the point/switch lever.

So, how is it done?
[Off-topic] stepie22:
Brain teaser/Puzzle

Hi all!!!

Today at school I was a bit bored so I made this puzzle for you guys.

"Key":

I / R / D / P / N / F / V / S / A / X / H / L / B / U / W / O / E / Y / J / Q / C / M / Z / G / T / K / (space) / ! / . / ? / -​

Hints:

Remember to number!

Learn the computers language!

Your Puzzle to solve:

1010 / 11 / 1001 / 11111 / 11 / 10001 / 111 / 10001 / 1100 / 10000 / 100 / 10001 / 10 / 1000 / 11011 / 1 / 1000 / 11011 / 1001 / 1111 / 10001 / 1000 / 10000 / 10110 / 10001 / 11100 / 11100 / 11100​

Take things litteraly + think logical!

Goodluck!!!

ps. if no-one gets it, I' post more hints in two days time (2 March)

If anyone has questions post and I'll try to answer them the best I can without giving away the answer.

Extra hints:
[Off-topic] Dblfstr:
:D

xda developers is awesome!!!
[Off-topic] stepie22:
Yay! Is wasn't that hard, was it? :p

1010 = 10 = X

11 = 3 = D

10010 = 9 = A

....
[Off-topic] stephj:
The puzzle in post #1167 is still running... Nearly four months............ so, meanwhile....

What are the positions of the bolts after one bolt has done one complete revolution around the other with the threads still engaged?
[Off-topic] I r o n M a n:
Okay, guys. Not sure if this is a puzzle but try it out-

Try rotating you right hand clockwise along the wrist while trying to move your right foot anti-clockwise.

It is impossible!

This is because the right side of our body is controlled by the left hemisphere and our brain is actually incapable of true multitasking. We just shift our attention from one task to the other.

Sent from my octa-core, badboy S4 using the new Tapatalk 4 Beta
[Off-topic] Seraz007:
^^^Nice one ^^^ got tangled in a mess :eek::D

Simple One Here

~~~~~~~~~~~~~~~~

a girl is inside a fort. there is an apple tree with unlimited apples

there are unlimited amount of gates through which the girl has to pass to reach outside

at each gate, there is a guard. He will ONLY allow the girl to pass if she gives the Guard half of the no. of apples she picked (if she chooses 10 she gives the guard 5)

~~the girl also takes ONLY ONE APPLE FROM THE AMOUNT SHE GAVE THE GUARD~~ (so she has 4 remaining>>5-1=4 duuhh :p)

the apples cannot be halved as in u cannot choose an odd no.

How many apples does the girl pick so that she will never run out of apples?

~~~~~~~~~~~~~~~~

Really easy ;)
[Off-topic] Dblfstr:
All of them (the whole tree).
[Off-topic] Ichigo:
I can do it perfectly. My brain is different from a normal human's brain I guess.
[Off-topic] I r o n M a n:
Not sure if I should feel astounded or trolled :p

Sent from my octa-core, badboy S4 using the new Tapatalk 4 Beta
[Off-topic] jRi0T68:
I had no problem doing it either. :shrug:

-No longer accidental, just Jeremy. F.cfb

[Guide] Headphone/Earphone Buying
[Off-topic] Seraz007:
umm a girl can't lift a tree soo and she can't carry all the at once..... u fail :p:p

there is a much easier answer :)
[Off-topic] Dblfstr:
Sure she can, in a world of unlimited gates, and an apple tree with unlimited apples. The girl clearly has unlimited strength. I don't think she even wants to go outside. She should just sit in the fort eating apples and throwing some at the first guard just for fun.
[Off-topic] Seraz007:
think logically she cant :p

much easier answer is there !! :p :confused:
[Off-topic] jRi0T68:
He makes a good point, though. There couldn't be unlimited apples.

-No longer accidental, just Jeremy. F.cfb

[Guide] Headphone/Earphone Buying
[Off-topic] Seraz007:
it's puzzling really :rolleyes::laugh:
[Off-topic] Dblfstr:
Therein lies my logic.
[Off-topic] Dblfstr:
Hard to put into words.

Train grabs B (T-B)

Returns B to where Train began. (B T A)

Train grabs A (T-A) and puts where B began. (A T)

Train travels through tunnel and grabs B. ( B-T)

Travel down to A and Hook B to A (T-B-A)

Travel back up to top. (A-B-T)

Drop A Trains starting point. (A B-T)

Drop B at A starting point.

Train travel down through tunnel CW

Tran grab A (A-T) and travel to to B starting point (T-A)

Train travel back to train starting point.

Cars are reversed.

---------- Post added at 08:24 AM ---------- Previous post was at 08:10 AM ----------

The moment about the axis of the top screw is inversely proportional the moment about the axis of the bottom screw. And the relative motion follows the same opposition. :P

My brain hurts, but it seems that as one rotates around the other it creates two opposing rotations. The topmost screw would travel in a counterclockwise motion (as far as its relationship to the other screw). Relative to the threads, this would be "unscrewing". The bottom most would be traveling in a clockwise motion. Relative to its threads, this would be "screwing in" The position of each screw relative the the other would not change.