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anonymous
 4 years ago
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 nonnegative 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 xaxis and yaxis at two distinct points. Can you find the sum of the intercepts ?
Good luck!
anonymous
 4 years ago
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 nonnegative 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 xaxis and yaxis at two distinct points. Can you find the sum of the intercepts ? Good luck!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1333293300537:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why did everyone leave? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now how do I get the number of positive integral solutions? I don't wanna use hit and trial method :/ We know \(y\le30\) and \(x\le40\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0HINT: Linear Diophantine equation.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0these problems would be unsolveable and fool will greet us April Fool's Day mmhmm <nods> that would be epic

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1) is very much solvable by almost anybody.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0both of \( (x, y) \) are not positive??? either x is negative or y is negative or both are negative?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0looks like n is greatest common divisor

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Diophantine sounds like organic chemistry. I'm googling it. 10?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol, he (Diophantus) is the sometimes called the father of algebra.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It took me time but the Linear Diophantine thing is really nice. Thanks.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0is the answer of 2 1??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01 is the not the right answer.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0must be infinity then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it's not infinity.

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.2There are 11 non negative integral solutions to the given diophantine equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well done @2bornot2b!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If other want, you may like to the post the solution too.

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.2Using euclidean reduction it can be shown that the integral solutions are of the form \(x=4n\) and \(y=3(n10)\) 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 (n10) 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
 4 years ago
Best ResponseYou've already chosen the best response.2And yes for n10=0, the value is true too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94: Ishaan you are very close.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i was thinking the same .. lol

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0but it seems 0 is nowhere close to 81*64

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0Isn't it obvious that if both \(x\) and \(y\) are nonpositive, then the largest possible value of \(n\) is \(0\)?

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0I'm talking about the second problem.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ... but both \( (x, y) \) are not positive. might mean both cannot be ++, but can be + or + ... just considering possibility

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is, but at first I presumed it was both x and y can't be nonpositive integral :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you'll read the first question you might understand the cause of my presumption :/

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0still, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh he probably meant "such that x and y are not both positive", meaning that only one of them can be positive.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0in that case .. Mr. Math must be right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We 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
 4 years ago
Best ResponseYou've already chosen the best response.0The answer of (2) is \(0\), unless something's wrong with the question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0FoolForMath for the 3rd question, Can you tell the value of r(3,2)?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath what do you say??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0FoolForMath WHAT SAY YOU? *This is more cinematic hehe*

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0What does "is tiled with" mean? (My English is failing me)!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm made up of, maybe

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0But why is \(r(2,1)=3\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ishaan, \( r(3,2)=18 \).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no you are trying to find n is r(3,2)=16?

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.0@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 nonpositive, then we can choose \(x>>> 0\ge y\) and then n would have no maximum.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0number of rectangles?? you mean there are lot's of rectangle???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not both postive ok i see r(3,2)=18

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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, 8164 = 17 you have to maximize this value.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0?? lost me on that one

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0x=inf, y=1, n = inf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since 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
 4 years ago
Best ResponseYou've already chosen the best response.0But 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
 4 years ago
Best ResponseYou've already chosen the best response.0sat, if x and y are nonnegative integers, then the the greatest integer that cannot be written in the form ax+by is ab − a − b. [assuming (a,b)=1 ]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0x=inf, y=1, n = inf what about this ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have an answer i like for the second to last one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whew. left the house, had a beer and thought about it a little more clearly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Congratz sat! :) You may like to post your solution for others.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not unless everyone is done

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright:) Third one is the hardest in my opinion. So well done agian! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhm for number 2, do you mean the greatest possible value of n for x<0 and y<0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is the answer to the rectangle one 5148?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@satellite73 Am I right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i got, yes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's been a while, so http://ideone.com/k7o4g My thought process was explained in the comments.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the complimentary problem, \(a^2\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94: \(a^2 \) is not the right answer.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath do you make these things up? or you have some referneces? coz you would be sooo smart if these are original ___

perl
 4 years ago
Best ResponseYou've already chosen the best response.0If the tangent the to the curve x√+y√=a√ at any point on cuts xaxis and yaxis at two distinct points. Can you find the sum of the intercepts ?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.0answer for (2) is it n=1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0#2 answer : 5039 Complimentary problem: \( a \)
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