solve this bianary message. convert this to text

01001000 01101001 00100000 01110100 01101000 01100101 01110010 01100101 00100000 00101100 00100000 01101001 00100000 01101000 01100001 01110110 01100101 00100000 01101101 01100001 01100100 01100101 00100000 01100001 00100000 01110011 01101001 01101101 01110000 01101100 01100101 00100000 01101101 01100101 01110011 01110011 01100001 01100111 01100101 00100000 01111001 01101111 01110101 00100000 01101000 01100001 01110110 01100101 00100000 01110100 01101111 00100000 01110011 01101111 01101100 01110110 01100101 00100000 01100001 01101110 01100100 00100000 01101001 01110100 00100000 01101001 01110011 00100000 01101001 01101110 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 00101100 00100000 01101001 01101110 00100000 01101111 01110010 01100100 01100101 01110010 00100000 01110100 01101111 00100000 01110011 01101111 01101100 01110110 01100101 00100000 01110100 01101000 01101001 01110011 00100000 01111001 01101111 01110101 00100000 01101101 01110101 01110011 01110100 00100000 01100011 01101111 01101110 01110110 01100101 01110010 01110100 00100000 01110100 01101000 01101001 01110011 00100000 01101001 01101110 01110100 01101111 00100000 01101100 01100101 01110100 01110100 01100101 01110010 01110011

Hi there , i have made a simple message you have to solve and it is in binary , in order to solve this you must convert this into letters

http://www.roubaixinteractive.com/PlayGround/Binary_Conversion/Binary_To_Text.asp

Stupid post. There are hundreds of encoding conventions used for such conversions.

Never mind that all you need to do is dump it in a simple coding tool (IDLE for Python) and it will do it for you in seconds.

If you are going to pose a puzzle at least make it somewhat interesting or relevant, like the "Cheryl's Birthday" question making the rounds.

Edited 4/16/2015 16:40:49

Cheryl have 2 friends, Albert and Bernard. She told them her birthday could be one of the following:

Mai 15th, 16th or 19th ;

June 17th or 18th ;

Julyt 14th or 16th;

August 14th, 15th or 17th.

She then tells Albert the right month, and Bernard the right day.

Albert then says: "I don't know your birthday, but i know Bernard don't know either"
to wich Bernard responds: "I didn't know her birthday then, but know i do know her birthday"
to wich Albert conclude: "Well now that you know, i also know!"

When is Cheryl's birthday?

Hmmm ... i wont tell you what type of code this is ...

110 155 155 056 056 056 040 167 145 154 154 040 144 157 156 145 040 061 062 065 143 150 062 060 071 040 054 040 171 157 165 040 141 162 145 040 157 156 145 040 163 164 145 160 040 143 154 157 163 145 162 040 164 157 040 146 151 156 151 163 150 151 156 147 040 164 150 151 163 040 160 165 172 172 154 145 040 054 040 141 156 144 040 163 157 040 151 163 040 145 166 145 162 171 157 156 145 040 145 154 163 145 040 054 040 151 164 040 163 145 145 155 163 040 164 157 040 155 145 040 164 150 141 164 040 151 040 154 151 153 145 040 063 040 144 151 147 151 164 040 156 165 155 142 145 162 163 054 040 141 156 144 040 151 164 040 151 163 040 141 160 160 141 162 145 156 164 154 171 040 151 163 040 143 157 144 145 040 054 040 163 157 154 166 145 040 164 150 151 163 040 160 165 172 172 154 145 040 141 156 144 040 164 150 145 040 156 145 170 164 040 163 164 145 160 040 141 167 141 151 164 163 040 171 157 165 056

also solve this

.. ..-. -.-- --- ..- .... .- ...- . ... --- .-.. ...- . -.. - .... . --- -.-. - .- .-.. .--. ..- --.. --.. .-.. . .- -... --- ...- . .. -- ... ..- .--. .-. .. ... . -.. --··-- .-- . .-.. .-.. ... --- .-.. ...- . - .... .. ... - .... . -. --··-- . -. .--- --- -.-- ·-·-·-

Edited 4/16/2015 16:46:04

125, I assume you solved it for yourself? I wasn't absolutely certain on the logic before reviewing the solution but I was pretty confident I had the right answer (I did). One of those where I knew the answer but couldn't fully explain why I knew it.

MMB, clearly octal which means your grouping is rather nonsensical. Once again it could be any encoding method (UTF-8, CP-1252) so it's a rather pointless exercise, as is the morse code.

Edited 4/16/2015 16:58:25

Yeah every type of code could be converted anyway so here is a challenge

there are 27 people in a party

10 people had coke
8 people had beer
4 people had wine
5 people had lemonade

Each type of drink has a dispenser which can hold 2 liter. and each cup can hold 500 ml , the cups need to be filled at its maximum capacity ( 500ml ) however the dispensers cant hold enough drinks to fill all 27 cups.

before the party he buys :cups , dispensers and drinks ( 500ml drinks cost half of 1L bottles)

Prices (£) :

Dispenser : £100
cups ( 10 per pack ) : £1.50
Coke ( 1 Liter ) : £6
Lemonade ( 1 Liter ) : £2.50
Wine and beer ( 1 Liter ) : £11.60 ( each )

How much money does he need to buy?


Im trying to find very difficult puzzles on the internet so please wait for a new one

Edited 4/16/2015 17:04:40

That's not hard to solve, just tedious. Need a pen and paper.

You have 12 coins , one of them is fake , the fake coin is identical in apperance but slightly differnt in weight . You have a balanced scale , the scale only tells you which side weighs more than the other.

theese are modern coins ,the fake coin is not nessacairaly lighter.

You have to use only the 12 coins , no extra scales no pencil marks on the scale.

what is the smaller number to times to use the scal to find the fake coin.

125, I assume you solved it for yourself? I wasn't absolutely certain on the logic before reviewing the solution but I was pretty confident I had the right answer (I did). One of those where I knew the answer but couldn't fully explain why I knew it.

Yes, i saw it posted on another forum, and i solved it. Its really not hard, it was meant to be solved by teens after all...

You have 12 coins , one of them is fake , the fake coin is identical in apperance but slightly differnt in weight . You have a balanced scale , the scale only tells you which side weighs more than the other.

theese are modern coins ,the fake coin is not nessacairaly lighter.

You have to use only the 12 coins , no extra scales no pencil marks on the scale.

what is the smaller number to times to use the scal to find the fake coin.

3 times most of the time, 4 times worst case scenario:


divide 12coins into 4 bags of 3 coins(A,B,C,D)

(1->)
do A vs B

if balanced do (2->) A vs C: if balanced then fake is in D if not then fake in C

if not balanced do (2->) A vs C: if balanced then the fake is in B if not then fake in A

so we used 2 times the scale to isolate 3 coins (x,y,z) and if fake is not in D then we know if fake is lighter or heavier (D worst case scenario)

(3->)
do x vs y

if balanced then fake is z

if not balanced then

if fake was in A,B,C we know if it is heavier or lighter so we know if it is x or y
if fake was in D then

do (4->)
x vs z : if balanced then fake is y if not balanced then fake is x

edit: i saw the answer, and using 3 times the scale is enough, so my method isn't the best one

Edited 4/16/2015 18:41:16

there are 1000 wine bottles and one of them is posioned there are 1000 guest coming to your palace tomorrow.

the posion shows no symtoms until death ( 24 hours) . You have 1000 slaves at disposal , they must determine the posioned bottle.

you have a handful of prisoners at sample the bottle to see which one is posioned

How many prisoners do you need?
How many prisoners will die ?

Tip: binary digit the wine bottles to make it easier.

Edited 4/16/2015 18:33:51

well if you have 1000 slave at your disposal then you just make each one of them taste 1 bottle. the one that died drank the wrong bottle...
actually you only need 999 slaves, and if no-one dies, the wrong bottle is the one that no-one tasted

sinep

solve that

Done.

Edited 4/16/2015 18:58:03

Yes that is correct , you only need 999 slaves to taste the bottle

What is the smallest amount of slaves?
How many people will die in the smallest number of slaves?

ok can you solve this ( its just a maths question , not really that hard )


A prsion has 240 prisoners

there are 5 sections North , Courtyard , East , West and South

The courtyard has 1/8 of prisoners at any given time ( except at 9pm which it is locked)
East and West has 30% prisoners each at any given time
North has 40% prisoners
What will south have


Solve this

Here are 20 3-digit numbers

381
672
901
666
592
167
835
198
156
923
549
187
266
701
492
926
887
165
111
667

1000 people pick their numbers

50 people choose each number

Then they are given a ticket having thier picked number writen on it.

Like a lottorey the numbers are written on balls and one number will drop out every round

The winning number will get £10,000

there are 156 rounds

What is The most you can win?
In total , what is the probability of you getting your ball out of all 156 rounds?
if there is a 1 in 20 chance of getting you numbered ball , what is the average number of balls will you get out of all the rounds.

Note : the winning balls are put back in

Edited 4/16/2015 19:44:39

1. 2. 3. 4. 5. 6

solve that

Test 1:
1-2-3-4 left, 5-6-7-8 right, 9-10-11-12 off

Case 1: If the scale says they are equal:
1-2-3 left, 9-10-11 right (measurement 2)
If the right is heavier, you know the fake is heavy. If lighter, you know the fake is light. If the same, you know 12 is the fake (possible two try solution). If the two were different:
9 left, 10 right, 11 off (measurement 3)
Because you know if the fake is heavier or lighter, you'll know which it is after this measurement.

Case 2: If the scale has one side heavier than the other:
4-5-6-7 left, 8-9-10-11 right (measurement 2)
Case 2.1 If the scales switch from test 1, you know the fake is 5, 6, or 7, and you also know if it's heavier or lighter. In that case, you put 5 on the left, 6 on the right, and 7 off the scale and you know the answer.
Case 2.2 If the scales don't change from the previous measurement, you know the fake is either 4 or 8. Measure 4 against any other coin. If the same, the fake is 8. If different, the fake is 4.
Case 2.3 If the scales have both sides as equal, you know the fake is 1, 2, or 3, and you know if the fake is heavier or lighter. Put 1 on the left, 2 on the right, and 3 off the scale and you know the answer.

What will south have?

negative 1/8?!

Are the winning numbers being replaced each round? Or, once it comes out, does it stay out?