|
I got this from a friend, it's pretty classic according to him
alpha and zetta had a flock of N sheeps
they sold them at N dollar each, for a total of N^2 dollars.
they put the dollars in a case.
alpha go to the case, and he took 10 dollars
zetta go to the case, and he took 10 dollars
...
... they alternate in taking moneys for awhile...
...
alpha go to the case, and he took 10 dollars
(a)zetta go to the case, and there is less than 10 dollars remaining, so he took the rest.
alpha made even with zetta by giving zetta some money
How much did zetta took from the case in step (a) ?
+ Show Spoiler [answer] +I think the answer is 6, I'll be sure if someone come up with 6 as well. Here's justification: Total number when modded by 20 has to be >10 since alpha took 10 then zetta has <10, which adds up to be >10. Last trip zetta took less than 10 dollars Total amount has to be even since alpha can make even so the last bit of money has to be 2,4,6,8 no square number can end with 2 or 8, so it is either 4 or 6. for a number to end in 4, and a square, we have N^2 = something4, then N has to end with 2. N = (10n + 2) N^2 = 100n^2 + 40n + 4, which, if modded by 20 has remainder of 4, which is not greater than 10. Thus zetta took a final of 6 dollars. edit: I missed a case but the end solution does not change, thanks asharai!!
Hello,
For your math puzzle #16. There is something missing in the solution. A Square number N^2 ending by 4 means that either N ends by 2 or N ends by 8.
You missed the 8 case, although it can be removed with same reasons as for the 2 case.
Asharai
So I'll just put it in here, if it ends with 8, then (10n+8)^2 = 100n^2 + 160n + 60 + 4 which also divides to less than 10 by 20.
I'll try to get people to rate my math blogs in terms of easiness
Please rate this blog 1 if you find it super hard, and rate it 5 if you find it super easy, or somewhere in between if it's not too hard or bad
GL HF! spoiler ur answer please! :D
 
|
I smell a conspiracy to get higher rated blogs
|
I got the answer but differently :S
+ Show Spoiler +30 dollars were taken. Zetta takes the remainder of less than 10. So a square number has to be between 30-40, which is 6
|
On November 29 2009 19:30 writer22816 wrote:
I smell a conspiracy to get higher rated blogs
but you see either way isn't it? Either the TL people think "Oh this is a shit blog rate it 1" or they think "LOL i'm so Smaaarrt~~"
by giving rating system this way they fuck themselves up either way.
In any regard i saw 2 votes for 5 which means this should be easy yet I only see 1 reply with real answer
|
Got the same solution with the same thinking  .
I guess we're right!
|
On November 29 2009 19:30 TonyL2 wrote:I got the answer but differently :S + Show Spoiler +30 dollars were taken. Zetta takes the remainder of less than 10. So a square number has to be between 30-40, which is 6
oh by the "..." i meant they alternate taking money for potentially many many times, so the total dollar can be larger than 30.
i cleared it up now
|
On November 29 2009 19:30 writer22816 wrote:
I smell a conspiracy to get higher rated blogs
I felt some strange force compelling me to find this puzzle quite surprisingly hard.
|
On November 29 2009 19:40 538 wrote:Show nested quote +On November 29 2009 19:30 writer22816 wrote:
I smell a conspiracy to get higher rated blogs I felt some strange force compeling me to find this puzzle quite surprisingly hard.
It's just awkwardly phrased. Once you realize what it's actually asking it's very basic math.
|
Hmm, I like this problem but I was confused by
+ Show Spoiler +the "must be even" conclusion, can't you split up an odd number of dollars evenly using cents??
After thinking about it though, there wouldn't be a single answer if you did that
|
On November 29 2009 19:44 [ZiNC]Ling wrote:Hmm, I like this problem but I was confused by + Show Spoiler +the "must be even" conclusion, can't you split up an odd number of dollars evenly using cents??
After thinking about it though, there wouldn't be a single answer if you did that
ah yeah dollars are unbreakable, didn't thhink people woiuld try to break dollars haha,
|
you confused me with the
"...
... they alternate in taking moneys for awhile...
..."
in your solution, you say they took 20xn dollar in this process but the text could say 10xn, cause it's not sure if the alternating happened even times
|
On November 29 2009 20:32 freelander wrote:
you confused me with the
"...
... they alternate in taking moneys for awhile...
..."
in your solution, you say they took 20xn dollar in this process but the text could say 10xn, cause it's not sure if the alternating happened even times
The relevant detail is that alpha was the first person to take $10 AND the last person to take $10, so there were $10 * odd# dollars initially. (Plus the less-than-$10 leftover that zetta took.)
|
6 it is for me as well
|
Oh that's cool, any perfect square of the form (10n + b) with n being odd and 0<b<10, needs b=6.
|
|
|
|
|
+ Show Spoiler +To satisfy the conditions N^2 has to be odd in the tens place.
That only happens when N ends in a 4 or 6.
Both of which give an N^2 that ends in 6.
( 0 could give odd tens, but no remainder)
Is there always a solution if they were taking some amount other than 10?
I think the answer would be 1 if they took 8 at a time.
|
Wow, I tried a good number of the puzzles and couldn't solve any X_X!! But it was fun mulling over the riddles. I was wondering why the puzzles suddenly end at this post. MOAR please!
|
|
|
|
|
|
EPT
