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FoolForMath
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!
Why did everyone leave? :(
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 \(x\le40\).
HINT: Linear Diophantine equation.
these problems would be unsolveable and fool will greet us April Fool's Day mmhmm <nods> that would be epic
(1) is very much solvable by almost anybody.
both of \( (x, y) \) are not positive??? either x is negative or y is negative or both are negative?
looks like n is greatest common divisor
Diophantine sounds like organic chemistry. I'm googling it. 10?
Lol, he (Diophantus) is the sometimes called the father of algebra.
It took me time but the Linear Diophantine thing is really nice. Thanks.
is the answer of 2 1??
1 is the not the right answer.
must be infinity then
No it's not infinity.
There are 11 non negative integral solutions to the given diophantine equation
Well done @2bornot2b!
If other want, you may like to the post the solution too.
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
And yes for n-10=0, the value is true too.
@Ishaan94: Ishaan you are very close.
i was thinking the same .. lol
but it seems 0 is nowhere close to 81*64
Isn't it obvious that if both \(x\) and \(y\) are non-positive, then the largest possible value of \(n\) is \(0\)?
I'm talking about the second problem.
yeah ... but both \( (x, y) \) are not positive. might mean both cannot be ++, but can be +- or -+ ... just considering possibility
It is, but at first I presumed it was both x and y can't be non-positive integral :/
If you'll read the first question you might understand the cause of my presumption :/
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
Oh he probably meant "such that x and y are not both positive", meaning that only one of them can be positive.
in that case .. Mr. Math must be right.
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|
The answer of (2) is \(0\), unless something's wrong with the question.
FoolForMath for the 3rd question, Can you tell the value of r(3,2)?
@FoolForMath what do you say??
FoolForMath WHAT SAY YOU? *This is more cinematic hehe*
What does "is tiled with" mean? (My English is failing me)!
But why is \(r(2,1)=3\)?
Ishaan, \( r(3,2)=18 \).
no you are trying to find n is r(3,2)=16?
@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.
number of rectangles?? you mean there are lot's of rectangle???
not both postive ok i see r(3,2)=18
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.
?? lost me on that one
x=inf, y=-1, n = inf
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
But as long as you have positive y intercept and x intercept, you will always have positive solutions for (x,y) subject to \(x,y \in \mathbb{R}\).
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 ]
x=inf, y=-1, n = inf what about this ??
i have an answer i like for the second to last one
whew. left the house, had a beer and thought about it a little more clearly
Congratz sat! :) You may like to post your solution for others.
not unless everyone is done
Alright:) Third one is the hardest in my opinion. So well done agian! :)
uhm for number 2, do you mean the greatest possible value of n for x<0 and y<0?
Is the answer to the rectangle one 5148?
@satellite73 Am I right?
that is what i got, yes
It's been a while, so http://ideone.com/k7o4g My thought process was explained in the comments.
For the complimentary problem, \(a^2\)?
@Ishaan94: \(a^2 \) is not the right answer.
@FoolForMath do you make these things up? or you have some referneces? coz you would be sooo smart if these are original -___-
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 ?
answer for (2) is it n=-1?
#2 answer : 5039 Complimentary problem: \( a \)