December 31st, 2016, 09:47 AM  #1 
Newbie Joined: Dec 2016 From: New Jersey Posts: 7 Thanks: 1 
Problem 1) A kcity is a k by k grid of positive integers so that every integer greater than 1 is the sum of one integer from the same row and one integer from the same column. Below is an example of a 3city: 1 1 2 2 3 1 6 4 1 Part 1: Make a 5city that has an integer that is greater than or equal to 150. Part 2: Show that for all k ≥ 1, the largest entry in a kcity is, at most, 3^ (k choose 2) Problem 2) Let there be the set R = {x + 1/x , where x ranges over all positive rational numbers} (A) Let A be a positive integer. Show that A is the sum of two elements of R if and only if A is the product of two elements of R. (B) Show that there are infinitely many positive integers A that cannot be written as the sum of two elements of R. (C) Show that there are infinitely many positive integers A that can be written as the sum of two elements of R. Problem 3) George, Harry, and Ethan are perfect logicians that always tell the truth. George decides to pose a puzzle to his friends: he tells them that the day of his birthday is at most the number of the month of his birthday. Then George announces that he will whisper the day of his birthday to Harry and the month of his birthday to Ethan, and he does exactly that. After George whispers to both of them, Harry thinks a bit, then says “Ethan cannot know what George's birthday is.” After that, Ethan thinks a bit, then says “Harry also cannot know what George's birthday is.” This exchange repeats, with Harry and Ethan speaking alternately and each saying the other can’t know George's birthday. However, at one point, Ethan instead announces “Harry and I can now know what George's birthday is. Interestingly, that was the longest conversation like that we could have possibly had before both figuring out George's birthday.” Find George's birthday. Last edited by skipjack; January 2nd, 2017 at 08:05 PM. 
December 31st, 2016, 10:35 AM  #2  
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637  Quote:
let $a = x y + \dfrac {1}{x y}$ let $b = \dfrac x y + \dfrac y x$ $A = x y + \dfrac {1}{x y} + \dfrac x y + \dfrac y x$ $A = \left(x + \dfrac 1 x\right)\left(y + \dfrac 1 y\right)$ clearly $x + \dfrac 1 x \in R,~y + \dfrac 1 y \in R$ so $A$ is the product of two elements of $R$ now assume $A = x y,~x,y \in R$ $A = (a + 1/a)(b + 1/b) = ab + \dfrac a b + \dfrac b a + \dfrac{1}{ab}$ $A = \left(a b + \dfrac {1}{ab}\right) + \left(\dfrac a b + \dfrac {1}{\frac b a}\right)$ $x = ab + \dfrac {1}{ab}$ $y =\dfrac a b + \dfrac {1}{\frac b a}$ $x, y \in R$ $A = x + y$ Last edited by skipjack; January 1st, 2017 at 02:28 PM.  
January 1st, 2017, 02:22 PM  #3 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  I don't see how you can justify the above "let"s.
Last edited by skipjack; January 1st, 2017 at 02:27 PM. 
January 1st, 2017, 02:26 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,204 Thanks: 1291 
They are definitions.

January 1st, 2017, 02:27 PM  #5 
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637  
January 1st, 2017, 02:45 PM  #6 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  In this part of the proof, the hypothesis is: "A is a positive integer such that $A = a + b,~a,b \in R$." So you can't define $a,b$. If you assert "let $a$ = ..." and "let $b$ = ...", it must be the case that the claimed forms for $a,b$ follow from the hypothesis. That's what I don't see. 
January 1st, 2017, 02:54 PM  #7 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  $a,b$ are arbitrary elements of R. So you can claim $a = u + 1/u$ and $b = v + 1/v$ for some positive rational numbers $u,v$, but I don't see how you can assert, without justification, that there exist positive rational numbers x,y such that $u = xy$ and $v =x/y$. 
January 1st, 2017, 02:58 PM  #8  
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637  Quote:
and form $a,b$ I'm not specifying the value of $A$, just that it's made up of the sum of two members of $R$  
January 1st, 2017, 03:05 PM  #9 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  You asserted a particular form for $a,b$. I agree that if $a,b$ have that form, then $a,b \in R$. Are you claiming that all pairs $a,b \in R$ can be put in that form? I don't see that.

January 1st, 2017, 03:07 PM  #10  
Senior Member Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637  Quote:
I see the problem. the assertion is false for probably most $A$, but it's true for all $A$ made up of the sum of two members of $R$ Last edited by romsek; January 1st, 2017 at 03:09 PM.  

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