## 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 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!

1. anonymous

|dw:1333293300537:dw|

2. anonymous

Why did everyone leave? :(

3. 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 $$x\le40$$.

4. anonymous

HINT: Linear Diophantine equation.

5. anonymous

these problems would be unsolveable and fool will greet us April Fool's Day mmhmm <nods> that would be epic

6. anonymous

(1) is very much solvable by almost anybody.

7. experimentX

both of $$(x, y)$$ are not positive??? either x is negative or y is negative or both are negative?

8. experimentX

looks like n is greatest common divisor

9. anonymous

Diophantine sounds like organic chemistry. I'm googling it. 10?

10. anonymous

Lol, he (Diophantus) is the sometimes called the father of algebra.

11. anonymous

It took me time but the Linear Diophantine thing is really nice. Thanks.

12. experimentX

is the answer of 2 1??

13. anonymous

1 is the not the right answer.

14. AravindG

......

15. experimentX

must be infinity then

16. anonymous

No it's not infinity.

17. 2bornot2b

There are 11 non negative integral solutions to the given diophantine equation

18. 2bornot2b

Am I right?

19. anonymous

Well done @2bornot2b!

20. 2bornot2b

Thanks!

21. anonymous

If other want, you may like to the post the solution too.

22. 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

23. 2bornot2b

And yes for n-10=0, the value is true too.

24. anonymous

For 2nd 81*64?

25. anonymous

@Ishaan94: Ishaan you are very close.

26. anonymous

81*64 - 1? lol

27. anonymous

haha, nopes :)

28. anonymous

0?

29. experimentX

i was thinking the same .. lol

30. experimentX

but it seems 0 is nowhere close to 81*64

31. anonymous

Yeah lol

32. Mr.Math

Isn't it obvious that if both $$x$$ and $$y$$ are non-positive, then the largest possible value of $$n$$ is $$0$$?

33. Mr.Math

I'm talking about the second problem.

34. experimentX

yeah ... but both $$(x, y)$$ are not positive. might mean both cannot be ++, but can be +- or -+ ... just considering possibility

35. anonymous

It is, but at first I presumed it was both x and y can't be non-positive integral :/

36. anonymous

If you'll read the first question you might understand the cause of my presumption :/

37. 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

38. Mr.Math

Oh he probably meant "such that x and y are not both positive", meaning that only one of them can be positive.

39. experimentX

in that case .. Mr. Math must be right.

40. 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|

41. Mr.Math

The answer of (2) is $$0$$, unless something's wrong with the question.

42. anonymous

FoolForMath for the 3rd question, Can you tell the value of r(3,2)?

43. experimentX

@FoolForMath what do you say??

44. anonymous

FoolForMath WHAT SAY YOU? *This is more cinematic hehe*

45. Mr.Math

What does "is tiled with" mean? (My English is failing me)!

46. anonymous

47. Mr.Math

Oh I see.

48. Mr.Math

But why is $$r(2,1)=3$$?

49. anonymous

Ishaan, $$r(3,2)=18$$.

50. anonymous

no you are trying to find n is r(3,2)=16?

51. 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.

52. experimentX

number of rectangles?? you mean there are lot's of rectangle???

53. anonymous

not both postive ok i see r(3,2)=18

54. 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.

55. anonymous

?? lost me on that one

56. experimentX

x=inf, y=-1, n = inf

57. 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

58. anonymous

But as long as you have positive y intercept and x intercept, you will always have positive solutions for (x,y) subject to $$x,y \in \mathbb{R}$$.

59. 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 ]

60. experimentX

61. anonymous

i have an answer i like for the second to last one

62. anonymous

whew. left the house, had a beer and thought about it a little more clearly

63. anonymous

Congratz sat! :) You may like to post your solution for others.

64. anonymous

not unless everyone is done

65. anonymous

Alright:) Third one is the hardest in my opinion. So well done agian! :)

66. anonymous

merci

67. anonymous

uhm for number 2, do you mean the greatest possible value of n for x<0 and y<0?

68. anonymous

Is the answer to the rectangle one 5148?

69. anonymous

@satellite73 Am I right?

70. anonymous

that is what i got, yes

71. anonymous

It's been a while, so http://ideone.com/k7o4g My thought process was explained in the comments.

72. anonymous

For the complimentary problem, $$a^2$$?

73. anonymous

@Ishaan94: $$a^2$$ is not the right answer.

74. anonymous

@FoolForMath do you make these things up? or you have some referneces? coz you would be sooo smart if these are original -___-

75. 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 ?

76. .Sam.

hmm

77. .Sam.

answer for (2) is it n=-1?

78. .Sam.

or it could be zero

79. anonymous

#2 answer : 5039 Complimentary problem: $$a$$

80. anonymous

Oh yeah, a

81. experimentX

explanation???