Consider the following equation:
S E N D + M O R E M O N E Y
S E N D + M O R E M O N E Y
Trivial Trials Testing Total Capacity of Thought
79 posts · 2006-07-21 02:08:49 to 2006-08-28 07:05:07
S E N D + M O R E M O N E Y
yourbuddy wrote:
Consider the Following equation:
S E N D
+ M O R E
M O N E Y
Each letter stands for a single digit. Find a coding that will make the above sum true. Good luck
sorry, messes up.Consider the following equation:
Each letter stands for a single digit. Find a coding that will make the above sum true.
S E N D + M O R E M O N E Y
Consider the Following equation:
S E N D
+ M O R E
M O N E Y
Each letter stands for a single digit. Find a coding that will make the above sum true. Good luck
k, here's what i got...
S=4 (4)(5)(1)(3) = 60
E=5 +(2)(6)(7)(5) = 420
N=1 --------------
D=3 (2)(6)(1)(5)(
= 480 
M=2
O=6
R=7
Y=8
Sry in advance if it doesnt come out right lol
S=4 (4)(5)(1)(3) = 60
E=5 +(2)(6)(7)(5) = 420
N=1 --------------
D=3 (2)(6)(1)(5)(
= 480 M=2
O=6
R=7
Y=8
Sry in advance if it doesnt come out right lol
Skill wrote:
k, here's what i got...yeah, sorta fixed
S=4 (4)(5)(1)(3) = 60
E=5 +(2)(6)(7)(5) = 420
N=1 --------------
D=3 (2)(6)(1)(5)(8 ) = 480
M=2
O=6
R=7
Y=8
Sry in advance if it doesnt come out right lol
Skill wrote:
S E N D
+ M O R E
--------------
M O N E Y
Skill wrote:Incorrect. Ok, I will explain it a little more. S E N D, M O R E = a number in the thousands, ie 1,234. Same type of thing for M O N E Y. No multiplication at all.k, here's what i got...yeah, sorta fixed
S=4 (4)(5)(1)(3) = 60
E=5 +(2)(6)(7)(5) = 420
N=1 --------------
D=3 (2)(6)(1)(5)(8 ) = 480
M=2
O=6
R=7
Y=8
Sry in advance if it doesnt come out right lol
S E N D
+ M O R E
--------------
M O N E Y
I don't think they're supposed to be multiplied, Skill.
O=0
M=1
Y=2
E=5
N=6
D=7
R=8
S=9
Send=9567
More=1085
Money=10652
9567
+1085
10652
dang was thinkin variables 
Kivo wrote:
Is correct. CongratsI don't think they're supposed to be multiplied, Skill.
O=0
M=1
Y=2
E=5
N=6
D=7
R=8
S=9
Send=9567
More=1085
Money=10652
9567
+1085
10652
After his long journey, Fingulfin was at a small airport in London, waiting for the direct flight to Shinjuku WC. While he was waiting he noticed an old acquaintance, the former King of Hazaa!
"How are you?" asked Fingulfin
"Not too good. There was a revolution, and I was exiled. I am now only the king of a very small hamlet"
"Well," said Fingulfin, "It is good to be the king of something, even a small hamlet."
"You would think. But there are only 66 adults in my hamlet, including me. And we all get the same salary, one nuxu per week. And everyone gets to vote at the weekly meetings, except me. I don't even get a vote!"
"Do you have any power at all?"
"All I am allowed to do at the weekly meetings is suggest a change in the salary distribution. I might suggest that I get six nuxu, that my friends BuddyTonto and Synakal each get thirty nuxu, and nobody else gets anything."
"And then what happens?"
"Then they vote on the change. The people whose salary would increase as a result of my proposal always vote 'Aye', and the people whose salary would decrease always vote 'Nay'."
"What about the people whose salary wouldn't change?"
"They don't vote. So if I suggest that I get six nuxu a week, and BuddyTonto and Synakal get thirty each, and nobody else gets anything, then the proposal would be defeated 63 - 2 (remember, I don't get a vote)."
"And what would it take for one of your proposals to pass?"
"It has never happened. For it to happen, there have to be more 'Aye' votes than 'Nay' votes.
[size="xx-small">[size="small"][face="arial,helvetica,sans-serif"]Fingulfin was silent after that. He realized that there was a way for the King to make a series of proposals that would result in a significant salary increase for himself. But he didn't want to say it, because it would impoverish many of the country's citizenry. What is the maximum salary the King could eventually obtain for himself, and how many weeks would it take for him
Aight - might be an easier way, but... After 7 weeks the king should have 63 nuxus.
The first week the king proposes that 33 citizens get a raise of 1 nuxu for a total of 2 each, and no one else including himself gets anything.
33 aye, 32 nay, 66 nuxus distributed.
The 2nd week the king proposes that 17 citizens get a raise of 1 nuxu for a total of 3 each, 16 citizens lose their pay, and the king gets 15.
17 aye, 16 nay, 66 nuxus distributed.
The 3rd week the king proposes that 9 citizens get a raise of 1 nuxu for a total of 4 each, 8 citizens lose their pay, and the king gets 30.
9 aye, 8 nay, 66 nuxus distributed.
The 4th week the king proposes that 5 citizens get a raise of 1 nuxu for a total of 5 each, 4 citizens lose their pay, and the king gets 41.
5 aye, 4 nay, 66 nuxus distributed.
The 5th week the king proposes that 3 citizens get a raise of 1 nuxu for a total of 6 each, 2 citizens lose their pay, and the king gets 48.
3 aye, 2 nay, 66 nuxus distributed.
The 6th week the king proposes that 2 citizens get a raise of 1 nuxu for a total of 7 each, 1 citizen loses pay, and the king gets 52.
2 aye, 1 nay, 66 nuxus distributed.
The 7th week the king proposes that 3 citizens, who are without any pay, get a raise of 1 nuxu each. And that the two citizens earning 7 nuxus lose their pay entirely. And the king gets 63.
3 aye, 2 nay, 66 nuxus distributed.
The first week the king proposes that 33 citizens get a raise of 1 nuxu for a total of 2 each, and no one else including himself gets anything.
33 aye, 32 nay, 66 nuxus distributed.
The 2nd week the king proposes that 17 citizens get a raise of 1 nuxu for a total of 3 each, 16 citizens lose their pay, and the king gets 15.
17 aye, 16 nay, 66 nuxus distributed.
The 3rd week the king proposes that 9 citizens get a raise of 1 nuxu for a total of 4 each, 8 citizens lose their pay, and the king gets 30.
9 aye, 8 nay, 66 nuxus distributed.
The 4th week the king proposes that 5 citizens get a raise of 1 nuxu for a total of 5 each, 4 citizens lose their pay, and the king gets 41.
5 aye, 4 nay, 66 nuxus distributed.
The 5th week the king proposes that 3 citizens get a raise of 1 nuxu for a total of 6 each, 2 citizens lose their pay, and the king gets 48.
3 aye, 2 nay, 66 nuxus distributed.
The 6th week the king proposes that 2 citizens get a raise of 1 nuxu for a total of 7 each, 1 citizen loses pay, and the king gets 52.
2 aye, 1 nay, 66 nuxus distributed.
The 7th week the king proposes that 3 citizens, who are without any pay, get a raise of 1 nuxu each. And that the two citizens earning 7 nuxus lose their pay entirely. And the king gets 63.
3 aye, 2 nay, 66 nuxus distributed.
the answer is 63, and it would take 7 weeks.
skeptic wrote:
Aight - might be an easier way, but... After 7 weeks the king should have 63 nuxus.Correct
The first week the king proposes that 33 citizens get a raise of 1 nuxu for a total of 2 each, and no one else including himself gets anything.
33 aye, 32 nay, 66 nuxus distributed.
The 2nd week the king proposes that 17 citizens get a raise of 1 nuxu for a total of 3 each, 16 citizens lose their pay, and the king gets 15.
17 aye, 16 nay, 66 nuxus distributed.
The 3rd week the king proposes that 9 citizens get a raise of 1 nuxu for a total of 4 each, 8 citizens lose their pay, and the king gets 30.
9 aye, 8 nay, 66 nuxus distributed.
The 4th week the king proposes that 5 citizens get a raise of 1 nuxu for a total of 5 each, 4 citizens lose their pay, and the king gets 41.
5 aye, 4 nay, 66 nuxus distributed.
The 5th week the king proposes that 3 citizens get a raise of 1 nuxu for a total of 6 each, 2 citizens lose their pay, and the king gets 48.
3 aye, 2 nay, 66 nuxus distributed.
The 6th week the king proposes that 2 citizens get a raise of 1 nuxu for a total of 7 each, 1 citizen loses pay, and the king gets 52.
2 aye, 1 nay, 66 nuxus distributed.
The 7th week the king proposes that 3 citizens, who are without any pay, get a raise of 1 nuxu each. And that the two citizens earning 7 nuxus lose their pay entirely. And the king gets 63.
3 aye, 2 nay, 66 nuxus distributed.
Fingulfin has a pendent with 9 red frags arrayed as so:
1 2 3
Object Object Patch
4 5 6
Object Object Object
7 8 9
Object Object Object
Fingulfin has a friend named Stot who has Coder tree still in its beta testing(OMFGWTFH4X). In this tree, there is an ablity named "alteration', and this ability allows Objects to turn into Patches, Patches to turn into Subroutines, and Subroutines into Objects. The problem is, there are still some bugs to be worked out with this ability. Instead of just changing one Red Frag, it changes all frags touching it. IE, if Stot used his ability on the frag in slot 6, the frags 5,6,9 would become patches, while the frag in slot 3 would become a subroutine!!! Stot wants to be able to stack all of his red frags together, so all of them have to be the same. What is the fewest times Stot has to use "Alteration"? And briefly explain how you dd it.
1 2 3
Object Object Patch
4 5 6
Object Object Object
7 8 9
Object Object Object
Fingulfin has a friend named Stot who has Coder tree still in its beta testing(OMFGWTFH4X). In this tree, there is an ablity named "alteration', and this ability allows Objects to turn into Patches, Patches to turn into Subroutines, and Subroutines into Objects. The problem is, there are still some bugs to be worked out with this ability. Instead of just changing one Red Frag, it changes all frags touching it. IE, if Stot used his ability on the frag in slot 6, the frags 5,6,9 would become patches, while the frag in slot 3 would become a subroutine!!! Stot wants to be able to stack all of his red frags together, so all of them have to be the same. What is the fewest times Stot has to use "Alteration"? And briefly explain how you dd it.
Well he could just stack them all as objects and use no alterations.
Um...nvm. I didn't see the patch there. Now I feel stupid. 
Stot needs 5 alterations to stack 9 Subs
1st alteration to slot 9 changes 6, 8, 9 to Patches
2nd alteration to slot 4 changes 1, 4, 5, 7 to Patches
3rd alteration to slot 3 changes 2 to Patch & 3, 6 to Subroutines
4th alteration to slot 8 changes 5, 7, 8, 9 to Subroutines
5th alteration to slot 1 changes 1, 2, 4 to Subroutines
1st alteration to slot 9 changes 6, 8, 9 to Patches
2nd alteration to slot 4 changes 1, 4, 5, 7 to Patches
3rd alteration to slot 3 changes 2 to Patch & 3, 6 to Subroutines
4th alteration to slot 8 changes 5, 7, 8, 9 to Subroutines
5th alteration to slot 1 changes 1, 2, 4 to Subroutines
skeptic wrote:
Stot needs 5 alterations to stack 9 SubsCorrect. Also, another 500k for the next person to post an alternate way of doing this.
1st alteration to slot 9 changes 6, 8, 9 to Patches
2nd alteration to slot 4 changes 1, 4, 5, 7 to Patches
3rd alteration to slot 3 changes 2 to Patch & 3, 6 to Subroutines
4th alteration to slot 8 changes 5, 7, 8, 9 to Subroutines
5th alteration to slot 1 changes 1, 2, 4 to Subroutines
You can rearrange the order of the changes but I haven't yet found a combination that doesn't use only slots 1, 3, 4, 8, & 9.
skeptic wrote:
You can rearrange the order of the changes but I haven't yet found a combination that doesn't use only slots 1, 3, 4, 8, & 9.This is all I wanted. I just wanted people test and try and just realize they can rearrange the order and it can still work. Not in an order though, are 5 arrangements of 1.3.4.8.9 that can work. You get 500k
lol - this has been fun.
I don't want to hijack your thread but I'll pass your $500,000 along & add $1,000,000 to the first person to correctly narrate the solution to the 3 Black Hats, 2 White Hats riddle. No diagrams or charts - you must be the first poster to correctly reply with a narrative description of the solution. If no one can answer this I will post the text of the riddle in approximately 24 hours.
Keep up the good work, Fin
I don't want to hijack your thread but I'll pass your $500,000 along & add $1,000,000 to the first person to correctly narrate the solution to the 3 Black Hats, 2 White Hats riddle. No diagrams or charts - you must be the first poster to correctly reply with a narrative description of the solution. If no one can answer this I will post the text of the riddle in approximately 24 hours.
Keep up the good work, Fin
I know the answer, but I won't post it yet. Since I check this thread more than anyone else does I am probably the first to look at your partake. I might be down for a few days, as I have to make some more problems. Probably only until the end of this week. Would be tomorrow, but classes just started again. I'll check later tonight and see if anyone has posted. And also, the 1.5 mil you are sending came from me most likely
First off - I like the Forehead Mark version better than the Hat version cause some wisenheimer always says that he can look up and see the color of a hat. But the riddle is commonly known as 3 Black Hats, 2 White Hats or maybe vice versa.
Three men compete. Each is given a single mark on the forehead. The marks are made with charcoal strips drawn from a lot of five. Three strips are black and two strips are white. The men will answer in order. The winner is the first to correctly state the color of his mark. Any man who guesses will be immediately zerged. The first man says he cannot state the color of his mark. The second man says he cannot state the color of his mark. The third man says he knows the color of his mark.
What is the color of the third man's mark, and how does he know?
Three men compete. Each is given a single mark on the forehead. The marks are made with charcoal strips drawn from a lot of five. Three strips are black and two strips are white. The men will answer in order. The winner is the first to correctly state the color of his mark. Any man who guesses will be immediately zerged. The first man says he cannot state the color of his mark. The second man says he cannot state the color of his mark. The third man says he knows the color of his mark.
What is the color of the third man's mark, and how does he know?
skeptic wrote:
First off - I like the Forehead Mark version better than the Hat version cause some wisenheimer always says that he can look up and see the color of a hat. But the riddle is commonly known as 3 Black Hats, 2 White Hats or maybe vice versa.In this, the third person cannot see either of the two correct? And the second person can only see the third person correct? Just want to make sure we are doing the same riddle.
Three men compete. Each is given a single mark on the forehead. The marks are made with charcoal strips drawn from a lot of five. Three strips are black and two strips are white. The men will answer in order. The winner is the first to correctly state the color of his mark. Any man who guesses will be immediately zerged. The first man says he cannot state the color of his mark. The second man says he cannot state the color of his mark. The third man says he knows the color of his mark.
What is the color of the third man's mark, and how does he know?
I'm presuming ceterus paribus, that the men have the same knowledge & can see each other.
Frell, no edit. Should be ceteris. Sorry.
Well, if they can all see eachother. The first 2 men each see a black and white stripe/hat. And the third person sees 2 white stripes, so he knows for sure he has a black stripe on his forehead.
But, I always thought the first could see both, the second could only the the third guy, and the third guy could see no-one.
Which means the first guy saw either 2 black or 1 black/1 white. If he saw two whites, he would know for a fact he had a black stripe, but b/c he did not know, it had to be one of the first two combinations.
If the third person had a white stripe, then the second person would know for a fact that he had a black stripe, b/c the first person did not know for a fact what color his stripe was. But b/c the second person did not know what color his stripe was, it means the third person had to be wearing a black stripe. This is why this third person knows for a fact he has a black stripe.
Again, you have to assume that this third person has the capable brain capacity to figure this out as well.
But, I always thought the first could see both, the second could only the the third guy, and the third guy could see no-one.
Which means the first guy saw either 2 black or 1 black/1 white. If he saw two whites, he would know for a fact he had a black stripe, but b/c he did not know, it had to be one of the first two combinations.
If the third person had a white stripe, then the second person would know for a fact that he had a black stripe, b/c the first person did not know for a fact what color his stripe was. But b/c the second person did not know what color his stripe was, it means the third person had to be wearing a black stripe. This is why this third person knows for a fact he has a black stripe.
Again, you have to assume that this third person has the capable brain capacity to figure this out as well.
Yes, close enough - or right spot on. You are correct that the second man only needs to see the third man and that the third man doesn't need to see the other two, however; the answer remains the same either way - I didn't want to say that before the riddle was solved.
Should any of the three men see two white marks then he knows he will win on his turn. Essentially, the first man can only win if he sees two white marks. If the first man loses, the second man can only win if he sees a white mark on the third man. And the third man can only win if the other two lose.
Should any of the three men see two white marks then he knows he will win on his turn. Essentially, the first man can only win if he sees two white marks. If the first man loses, the second man can only win if he sees a white mark on the third man. And the third man can only win if the other two lose.
i got a headache
darkhaze wrote:
i got a headacheOn;y because you're inferior