Riddle me this!

[Izzo]

Originally posted by: dnetmhz
What row of numbers comes next?
1
11
21
1211
111221
312211
13112221

the answer will have probably been revealed by the time my slowass computer submits the reply, but here goes: 1113213211

 

[DnetMHZ]

Originally posted by: Izzo

Originally posted by: dnetmhz
What row of numbers comes next?
1
11
21
1211
111221
312211
13112221

the answer will have probably been revealed by the time my slowass computer submits the reply, but here goes: 1113213211

you got it!

 

[MegaloManiaK]

Originally posted by: dnetmhz

Originally posted by: Izzo

Originally posted by: dnetmhz
What row of numbers comes next?
1
11
21
1211
111221
312211
13112221

the answer will have probably been revealed by the time my slowass computer submits the reply, but here goes: 1113213211

you got it!

I dont get it. Is it a pattern?

 

[DnetMHZ]

Originally posted by: MegaloManiaK

Originally posted by: dnetmhz

Originally posted by: Izzo

Originally posted by: dnetmhz
What row of numbers comes next?
1
11
21
1211
111221
312211
13112221

the answer will have probably been revealed by the time my slowass computer submits the reply, but here goes: 1113213211

you got it!

I dont get it. Is it a pattern?

no..each line describes the one above..

 

[CaseTragedy]

Originally posted by: dnetmhz

Originally posted by: MegaloManiaK

Originally posted by: dnetmhz
What row of numbers comes next?

1
11
21
1211
111221
312211
13112221

The eighth row?

You said what row comes next, not what sequence of numbers?

no taking the easy way out mister!

pattern riddles makes head hurt...

 

[DnetMHZ]

a better explaination:

the last line was: 13112221

the answer was: 1113213211

which works out to

one 1 one 3 two 1's three 2's one 1

 

[hawkeye81x]

I see why I.Q. is so low...
*applause*

 

[DnetMHZ]

ok, here is an easy one

I have holes on the top and bottom.
I have holes on my left and on my right.
And I have holes in the middle, yet I still hold water.
What am I?

 

[hoyaguru]

Originally posted by: dnetmhz
ok, here is an easy one

I have holes on the top and bottom.
I have holes on my left and on my right.
And I have holes in the middle, yet I still hold water.
What am I?

A sponge

 

[MegaloManiaK]

Originally posted by: dnetmhz
ok, here is an easy one

I have holes on the top and bottom.
I have holes on my left and on my right.
And I have holes in the middle, yet I still hold water.
What am I?

Heh, the earth

or a person, althought that gets a little messy

 

[DnetMHZ]

Originally posted by: hoyaguru

Originally posted by: dnetmhz
ok, here is an easy one

I have holes on the top and bottom.
I have holes on my left and on my right.
And I have holes in the middle, yet I still hold water.
What am I?

A sponge

that's it



Sally and her younger brother were fighting. Their mother was tired of the fighting, and
decided to punish them by making them stand on the same piece of newspaper in such a way that they couldn't touch each other. How did she accomplish this?

 

[Spooner]

If EELS + MARK + BEST + WARY = EASY
What does HELP + BARK + WARD + LEAD equal?

 

[Spooner]

Sally and her younger brother were fighting. Their mother was tired of the fighting, and
decided to punish them by making them stand on the same piece of newspaper in such a way that they couldn't touch each other. How did she accomplish this?

She slid it under a door

 

[DnetMHZ]

Originally posted by: Spooner
If EELS + MARK + BEST + WARY = EASY
What does HELP + BARK + WARD + LEAD equal?

HARD

 

[DnetMHZ]

Originally posted by: Spooner

Sally and her younger brother were fighting. Their mother was tired of the fighting, and
decided to punish them by making them stand on the same piece of newspaper in such a way that they couldn't touch each other. How did she accomplish this?

She slid it under a door

yes

 

[Spooner]

If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

 

[DnetMHZ]

Originally posted by: Spooner
If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

20? I'm assuming it goes by the # of letters in the name?

 

[Spooner]

Originally posted by: dnetmhz

Originally posted by: Spooner
If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

20? I'm assuming it goes by the # of letters in the name?

close, but wrong logic

your logic doesn't even fit the scenario

 

[MegaloManiaK]

Originally posted by: Spooner

Originally posted by: dnetmhz

Originally posted by: Spooner
If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

20? I'm assuming it goes by the # of letters in the name?

close, but wrong logic

your logic doesn't even fit the scenario

15?

 

[DnetMHZ]

Originally posted by: Spooner

Originally posted by: dnetmhz

Originally posted by: Spooner
If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

20? I'm assuming it goes by the # of letters in the name?

close, but wrong logic

your logic doesn't even fit the scenario

it could given ranges of lengths related to points

 

[Spooner]

15 is right, it's based on syllables.


You have two hourglasses--a 4-minute glass and a 7-minute glass.
You want to measure 9 minutes. How do you do it?

 

[MegaloManiaK]

Originally posted by: dnetmhz

Originally posted by: Spooner

Originally posted by: dnetmhz

Originally posted by: Spooner
If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but
Richard is 10, how much is Jennifer by the same system?

20? I'm assuming it goes by the # of letters in the name?

close, but wrong logic

your logic doesn't even fit the scenario

it could given ranges of lengths related to points

I think its 5 points per syllable = 15.

 

[Goth]

Originally posted by: Spooner
15 is right, it's based on syllables.


You have two hourglasses--a 4-minute glass and a 7-minute glass.
You want to measure 9 minutes. How do you do it?

Start both at the same time. Once the 4 minute glass is empty, there are 3 minutes remaining in the 7 minute one. Repeat two additional times.

Probably not the answer, but that's a way to get 9 minutes

 

[DnetMHZ]

Originally posted by: Goth

Originally posted by: Spooner
15 is right, it's based on syllables.


You have two hourglasses--a 4-minute glass and a 7-minute glass.
You want to measure 9 minutes. How do you do it?

Start both at the same time. Once the 4 minute glass is empty, there are 3 minutes remaining in the 7 minute one. Repeat two additional times.

Probably not the answer, but that's a way to get 9 minutes

you didn't account for getting the 7 minute glass reset.

 

[Goth]

Originally posted by: dnetmhz

Originally posted by: Goth

Originally posted by: Spooner
15 is right, it's based on syllables.


You have two hourglasses--a 4-minute glass and a 7-minute glass.
You want to measure 9 minutes. How do you do it?

Start both at the same time. Once the 4 minute glass is empty, there are 3 minutes remaining in the 7 minute one. Repeat two additional times.

Probably not the answer, but that's a way to get 9 minutes

you didn't account for getting the 7 minute glass reset.

I know, but the original question didn't say to measure a total of 9 sequential minutes. However, that's probably what needs to be measured and my answer is wrong, hence my above.