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dw5chaosfan

Answer that riddle

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ABCD= 2368

take away the + sign?

same amount of 1's and 2's

#1 is correct! The other two, wrong.

1) above

2) flip it over

3) number of ones add up to 4 two's also add up to 4

#2 is correct, but #3 is wrong.

Hint for #3: It's like no other equation when it comes to this.

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Erm there's only one way to arrange the numbers?

You can arrange them like 1+12=2+11 or 12+1=2+11 and other ways so nope.

Another hint: Rearranging

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11+2=12+1

Eleven+Two=Twelve+One

*thinks*

I dont understand DX

You are on the right track with that equation with the words. Now just think deeper :)

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I think I get it ... :unsure:

11+2=12+1 ---> Eleven+Two=Twelve+One

If you re-arrange the letters..

Twelve+One also spell out Eleven+Two and vice-versa

Nice job! :D

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Nice job! :D

Thanks, but it's only because you gave out some hints, otherwise, I probably couldn't solve riddles for my life, but I like em' cause' they're challenging..

I have some more riddles, probably won't be very long until you guys figure em' out though..

1.) 'What belongs to you, but is used more often by your friends?'

2.)'I am taken from a mine and shut up in a wooden case from which I am never released and yet I am used by almost everybody. What am I?'

3.)'If a fire hydrant has H2O inside what does it have on the outside?'

4.)'A doctor and a bus driver are both in love with the same woman an attractive girl named Sarah. The bus driver had to go on a long bustrip that would last a week. Before he left he gave Sarah seven apples. Why?'

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1) Your name.

2)Lead. (For Pencils)

3)K9P. (Canine p**)

4)To keep the doctor away while he was gone. (An apple a day...)

Great job~! Got ALL of them right!

*sigh* I'm just not good at giving riddles :(

Anyone else want to give it a shot~? ^_^

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well I'll post next riddle

it's pretty hard if you don't think

The four people in this puzzle all competed in different classes of dog agility at a recent competition. The competitions all required the dogs to run over jumps, through tunnels and various other obstacles in as quicker time as possible. Each had a different result - one came first, one third, one fourth and one ninth. All four dogs were each of a different breed.

Can you work out who handled which dog, at what level each competed, the place each finished in and the breed of each dog?

1. If Tiff finished first then Terry finished fourth.

2. If Terry finished fourth then Jago is a collie otherwise Jago is not a collie.

3. If Jane competed in the Senior class then she finished third.

4. If Jane competed in Novice then she finished fourth.

5. The dog that finished ninth was an alsatian. This was either Jago, in which case Jago competed in the Elementary class, or this was Kelly, in which case Terry handled Kelly.

6. Mark won Starters.

7. If Mark's dog is called Patti then Patti is a labrador otherwise Patti is a collie.

8. Ruth's dog is called Jago.

9. If Jago finished fourth then she competed in the Novice class otherwise she competed in the Senior class.

10. If Patti finished first then Terry's dog is an alsatian otherwise Terry's dog is a collie.

11. If Jane's dog is a doberman then Jane finished fourth otherwise Jane finished third.

Handler's Names: Jane, Mark, Ruth and Terry

Dog's Names: Tiff, Patti, Jago and Kelly

Breed: Alsatian, Collie, Labrador and Doberman

Level: Starters, Elementary, Novice or Senior

Answer should be like this

Dogs Handler's class place breeds

1 -------- ----- ----- ------

IN place of"---" fill them up four of them.

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Alright let's see if I understood your rules and how to post it.

1st) Mark/Patti/Labrador/Starter

3rd) Jane/Tiff/Collie/Senior

4th) Ruth/Jago/Doberman/Novice

9th) Terry/Kelly/Alsatian/Elementary

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Alright let's see if I understood your rules and how to post it.

1st) Mark/Patti/Labrador/Starter

3rd) Jane/Tiff/Collie/Senior

4th) Ruth/Jago/Doberman/Novice

9th) Terry/Kelly/Alsatian/Elementary

correct time for next riddle

A woman had three old coins - a silver dollar, a quarter, and a dime. Each coin was a little battered and had a piece missing. She found that exactly the same fraction had broken away from each coin.

What fraction of each was missing if the value of the remaining bits of coins was now exactly one dollar in total?

For this puzzle it can be assumed that ½ of a coin is worth ½ of its value.

Hint:-A dollar is 100 cents, a quarter is 25 cents and a dime is 10 cents.

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Oooh is this math? :P

---------------------------------

Let a be the fraction broken off.

100 x a + 25 x a + 10 x a = 100

so

a (100 + 25 + 10) = 100

a = 100/ (135)

a = 20/27

Is this the answer? :P

----------------

Edits: I just realised the question was...

"What is left is one dollar" instead of what's broken off... D:

-------- Okay rewrite answer --------

Let a be the fraction left. :V

So, the fraction broken off = 1 - a = 7/27

------------------------------------

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Oooh is this math? :P

---------------------------------

Let a be the fraction broken off.

100 x a + 25 x a + 10 x a = 100

so

a (100 + 25 + 10) = 100

a = 100/ (135)

a = 20/27

Is this the answer? :P

----------------

Edits: I just realised the question was...

"What is left is one dollar" instead of what's broken off... D:

-------- Okay rewrite answer --------

Let a be the fraction left. :V

So, the fraction broken off = 1 - a = 7/27

------------------------------------

awesomely done 2'nd one is Correct

Time for next riddle

What value of * makes the following correct

*/* - */2 + */4 =*/12

IN here "/" doesn't necessarily mean divide

Hint :- The answer is a whole number less than 10.

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12 - 12a/2 + 12a/4 = a

12 - 6a + 3a = a

12 = 6a - 3a +a

12=4a

a = 3

*= 3?

This looks like an equation, so erm pardon me if I'm wrong...

That's what I thought as well. But then aika said that it the slashs don't mean divide so I have no idea if this is right. :unsure:

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