anonymous
  • anonymous
Fool's problems of the day, [** EDIT: Complimentary problem added **] On the April fools' day I give you three cute problems on number theory/Combinatorics: \((1) \) Find the number of non-negative integral solutions of \(3x+4y=120 \) [Solved: @2bornot2b ] \( (2) \) If \(81x + 64y = n\) find the greatest possible of \( n \) such that both \( (x, y) \) are not positive. \( (3) \) Let \(n\) and \( m\) be positive integers. An \( n \times  m \) rectangle is tiled with unit squares. Let \(r(n,m) \) denote the number of rectangles formed by the edge of these unit squares. Thus, for example, \(r(2, 1) = 3\). Can you find \( r(11, 12) \)?. [ Solved by @satellite73] PS: The problems are arranged in an increasing order of difficulty. (However, (2) could seem very very easy to a number theorist!) ** Complimentary for those who feels these are very hard:** If the tangent the to the curve \(\sqrt{x} + \sqrt{y} = \sqrt{a} \) at any point on cuts x-axis and y-axis at two distinct points. Can you find the sum of the intercepts ? Good luck!
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1333293300537:dw|
anonymous
  • anonymous
Why did everyone leave? :(
anonymous
  • anonymous
Now how do I get the number of positive integral solutions? I don't wanna use hit and trial method :/ We know \(y\le30\) and \(x\le40\).

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anonymous
  • anonymous
HINT: Linear Diophantine equation.
lgbasallote
  • lgbasallote
these problems would be unsolveable and fool will greet us April Fool's Day mmhmm that would be epic
anonymous
  • anonymous
(1) is very much solvable by almost anybody.
experimentX
  • experimentX
both of \( (x, y) \) are not positive??? either x is negative or y is negative or both are negative?
experimentX
  • experimentX
looks like n is greatest common divisor
anonymous
  • anonymous
Diophantine sounds like organic chemistry. I'm googling it. 10?
anonymous
  • anonymous
Lol, he (Diophantus) is the sometimes called the father of algebra.
anonymous
  • anonymous
It took me time but the Linear Diophantine thing is really nice. Thanks.
experimentX
  • experimentX
is the answer of 2 1??
anonymous
  • anonymous
1 is the not the right answer.
AravindG
  • AravindG
......
experimentX
  • experimentX
must be infinity then
anonymous
  • anonymous
No it's not infinity.
2bornot2b
  • 2bornot2b
There are 11 non negative integral solutions to the given diophantine equation
2bornot2b
  • 2bornot2b
Am I right?
anonymous
  • anonymous
Well done @2bornot2b!
2bornot2b
  • 2bornot2b
Thanks!
anonymous
  • anonymous
If other want, you may like to the post the solution too.
2bornot2b
  • 2bornot2b
Using euclidean reduction it can be shown that the integral solutions are of the form \(x=4n\) and \(y=-3(n-10)\) where n is any integer. One can also use the formula to solve diophantine equation or any other method. Now for positive n, 4n is always positive, so we can't take into account any negative n. Again the expression (n-10) must be negative so as to have positive y, which is possible for n=0,1,2,3....,10. So we have 11 solutions
2bornot2b
  • 2bornot2b
And yes for n-10=0, the value is true too.
anonymous
  • anonymous
For 2nd 81*64?
anonymous
  • anonymous
@Ishaan94: Ishaan you are very close.
anonymous
  • anonymous
81*64 - 1? lol
anonymous
  • anonymous
haha, nopes :)
anonymous
  • anonymous
0?
experimentX
  • experimentX
i was thinking the same .. lol
experimentX
  • experimentX
but it seems 0 is nowhere close to 81*64
anonymous
  • anonymous
Yeah lol
Mr.Math
  • Mr.Math
Isn't it obvious that if both \(x\) and \(y\) are non-positive, then the largest possible value of \(n\) is \(0\)?
Mr.Math
  • Mr.Math
I'm talking about the second problem.
experimentX
  • experimentX
yeah ... but both \( (x, y) \) are not positive. might mean both cannot be ++, but can be +- or -+ ... just considering possibility
anonymous
  • anonymous
It is, but at first I presumed it was both x and y can't be non-positive integral :/
anonymous
  • anonymous
If you'll read the first question you might understand the cause of my presumption :/
experimentX
  • experimentX
still, considering this http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations i still think 0 is the answer, since 81and 64 do not have any common divisor
Mr.Math
  • Mr.Math
Oh he probably meant "such that x and y are not both positive", meaning that only one of them can be positive.
experimentX
  • experimentX
in that case .. Mr. Math must be right.
anonymous
  • anonymous
We need to minimize the indicated portion, which is only possible if \(n\le 0\), and 0 is the largest as the rest of the values are negative. |dw:1333297533937:dw|
Mr.Math
  • Mr.Math
The answer of (2) is \(0\), unless something's wrong with the question.
anonymous
  • anonymous
FoolForMath for the 3rd question, Can you tell the value of r(3,2)?
experimentX
  • experimentX
@FoolForMath what do you say??
anonymous
  • anonymous
FoolForMath WHAT SAY YOU? *This is more cinematic hehe*
Mr.Math
  • Mr.Math
What does "is tiled with" mean? (My English is failing me)!
anonymous
  • anonymous
Hmm made up of, maybe
Mr.Math
  • Mr.Math
Oh I see.
Mr.Math
  • Mr.Math
But why is \(r(2,1)=3\)?
anonymous
  • anonymous
Ishaan, \( r(3,2)=18 \).
anonymous
  • anonymous
no you are trying to find n is r(3,2)=16?
Mr.Math
  • Mr.Math
@satellite, if both \(x\) and \(y\) are not positive then obviously the maximum value of \(n\) is \(0\). If only one of them has to be non-positive, then we can choose \(x>>> 0\ge y\) and then n would have no maximum.
experimentX
  • experimentX
number of rectangles?? you mean there are lot's of rectangle???
anonymous
  • anonymous
not both postive ok i see r(3,2)=18
anonymous
  • anonymous
I think both \( x \) and \( y\) are not positive means either of them can be positive and both of them can be negative. x=1, y=-1 then, 81-64 = 17 you have to maximize this value.
anonymous
  • anonymous
?? lost me on that one
experimentX
  • experimentX
x=inf, y=-1, n = inf
anonymous
  • anonymous
since gcd(81,64)=1 we can solve \[81x+64y=1\] for x and y and so can solve \[81x+64y=n\] for any n
anonymous
  • anonymous
But as long as you have positive y intercept and x intercept, you will always have positive solutions for (x,y) subject to \(x,y \in \mathbb{R}\).
anonymous
  • anonymous
sat, if x and y are non-negative integers, then the the greatest integer that cannot be written in the form ax+by is ab − a − b. [assuming (a,b)=1 ]
experimentX
  • experimentX
x=inf, y=-1, n = inf what about this ??
anonymous
  • anonymous
i have an answer i like for the second to last one
anonymous
  • anonymous
whew. left the house, had a beer and thought about it a little more clearly
anonymous
  • anonymous
Congratz sat! :) You may like to post your solution for others.
anonymous
  • anonymous
not unless everyone is done
anonymous
  • anonymous
Alright:) Third one is the hardest in my opinion. So well done agian! :)
anonymous
  • anonymous
merci
anonymous
  • anonymous
uhm for number 2, do you mean the greatest possible value of n for x<0 and y<0?
anonymous
  • anonymous
Is the answer to the rectangle one 5148?
anonymous
  • anonymous
@satellite73 Am I right?
anonymous
  • anonymous
that is what i got, yes
anonymous
  • anonymous
It's been a while, so http://ideone.com/k7o4g My thought process was explained in the comments.
anonymous
  • anonymous
For the complimentary problem, \(a^2\)?
anonymous
  • anonymous
@Ishaan94: \(a^2 \) is not the right answer.
lgbasallote
  • lgbasallote
@FoolForMath do you make these things up? or you have some referneces? coz you would be sooo smart if these are original -___-
perl
  • perl
If the tangent the to the curve x√+y√=a√ at any point on cuts x-axis and y-axis at two distinct points. Can you find the sum of the intercepts ?
.Sam.
  • .Sam.
hmm
.Sam.
  • .Sam.
answer for (2) is it n=-1?
.Sam.
  • .Sam.
or it could be zero
anonymous
  • anonymous
#2 answer : 5039 Complimentary problem: \( a \)
anonymous
  • anonymous
Oh yeah, a
experimentX
  • experimentX
explanation???

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