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NorCal Melee Power Rankings - Summer '15 Update - In Sickness and In Filth

pockyD

Smash Legend
Joined
Jul 21, 2006
Messages
11,926
Location
San Francisco, CA
you should seed so i dont get ***** next time :p

Not to whine, but I've hit either Jeff or Zhu every time i've been back for a tourney in my 2nd round of winners (if there were no pools)
no johns! get better!

i've hit shroomed and jeff a lot recently too! oh well, you should be appreciating the opportunity to play the best! they aren't untouchable (and i don't mean that in a rah-rah get'em! way... i mean they are legitimately beatable for a player of your ilk)

it's random anyway
 

Shroomed

Smash Master
Joined
Feb 5, 2008
Messages
4,793
Location
Santa Cruz
no johns! get better!

i've hit shroomed and jeff a lot recently too! oh well, you should be appreciating the opportunity to play the best! they aren't untouchable (and i don't mean that in a rah-rah get'em! way... i mean they are legitimately beatable for a player of your ilk)

it's random anyway
So u call Jeff by name but I'm still shroomed?
 

choknater

Smash Obsessed
Joined
Dec 25, 2002
Messages
27,296
Location
Modesto, CA
NNID
choknater
ill try to do well at cgc so that i can get ranked

heres my record so far since gsg

wins -
boback, phil twice, tang, hmw, mexican twice

losses -
pewpewu, replicate twice, mew2king

see u guys at cgc!

not really trippin too much if i dont get ranked hehe i just think itd be cool!
 

WassupJB

Smash Lord
Joined
Oct 28, 2007
Messages
1,452
Location
NorCal
i don't type this much at all but LMFAO! dre's post then dunskies picture had me rooooooolllllllllling. omg i'm dying. this shouldn't be that funny.

don't take that dajuan.
 

pockyD

Smash Legend
Joined
Jul 21, 2006
Messages
11,926
Location
San Francisco, CA
somewhat 'fun' number puzzle

A teacher says: I'm thinking of two natural numbers bigger than 1. Try to guess what they are.
The first student knows their product and the other one knows their sum.
First: I do not know the sum.
Second: I knew that. The sum is less than 14.
First: I knew that. However, now I know the numbers.
Second: And so do I.
What were the numbers?
 

HyugaRicdeau

Baller/Shot-caller
Joined
Jun 4, 2003
Messages
3,883
Location
Portland, OR
Slippi.gg
DRZ#283
I'm fairly certain the numbers are
2 and 9.
It can't be 6 and 7 because it's possible to factor 6*7=42 into 2 and 21, which sum to 23, so if that really WERE the product, there's no way the first student could have KNOWN the sum were less than 14.

My reasoning:
The first student knows the product but not the sum, and also knows that the sum is less than 14. The fact that the sum is less than 14 is not a separate piece of info he is given, it's something he deduced with knowledge of the product: whatever product he knows, he must have been able to use it to deduce that there are multiple ways to factor it into 2 numbers whose sum is less than 14, and also that there are -no- ways to factor it into 2 numbers whose sum is 14 or more. This eliminates all products except 12, 18, and 20.

The second student knows the sum but not the product. He also knows that it is impossible to deduce the sum from the product. This means that whatever sum he knows, he is able to use it to deduce that knowledge of the product (which he doesn't know) does not lead to knowledge of the sum. This means that the sum must be such that ALL ways of splitting it into two numbers give numbers whose product is such that it has at least two ways of factoring it. There is only one such number, which is 11. For a while I was hung up on the fact that 4*7=28 and the only alternate way of factoring it is 2 and 14 which sum to 16. But the second student doesn't know that the first student knows the sum is less than 14, so this is OK.

There is only one pair of numbers that fits into both categories.

P.S. Hi everyone from Krakow. It is raining now so we can't do much, which is why I had time to solve the problem and type up the answer.
 

pockyD

Smash Legend
Joined
Jul 21, 2006
Messages
11,926
Location
San Francisco, CA
Also @ Sheridan 16 is a potential product as well, with 2 * 8 and 4 * 4, but the question is somewhat vague about whether the 2 numbers are actually different [the person who told me the riddle did specify they could be the same] (and it doesn't affect the answer anyway, since you can pare it away using the same logic you use for the other potential products)
 

Leruinoir

Smash Rookie
Joined
Sep 1, 2010
Messages
3
I live in berkeley, I moved here from hawaii, AND I WANT SOMEONE TO HOST A SMASH FEST IN BERKELEY. I have no friends who smash here and would like to make some!! email me? MKDamaso23@gmail.com
 
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