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Creative Problem Solving
I'm taking this course right now and it seems like the kind of thing a lot of people here would enjoy. It's basically a course full of "make four liters using only a three-liter jug and a five-liter jug"-type problems, although those problems are, of course, the easiest to solve.
Try some: - Let a, b and c be positive numbers. Prove that (a/b) + (b/c) + (c/a) >= 3. - For which real numbers c is there a straight line that intersects the curve y = (x^4) + 9(x^3) + c(x^2) + 9x + 4 at four distinct points? - How many arithmetic problems of the following form are possible? You must use each of the digits 1 through 9 in numerical order from left to right, and you can use any combination of the + and * symbols as you like provided your syntax is still correct. Proper examples include 123+45*6789, 1+2+3+4+5*6+7+89 and 123456789, while 12+3*+456++78*9 does not count. Now obviously I'm not going to ask any questions where I don't already know the answer, so don't bother accusing me of not doing my own homework. >_> |
Re: Creative Problem Solving
first one is AM-GM
second one "straight line" = ignore the linear and constant terms (x - a)(x - b)(x - c)(x - d) = x^{4} - (a + b + c + d) x^{3} + (ab + ac + ad + bc + bd + cd) x^{2} + useless **** basically a + b + c + d = -9 and we want to know the range of C = (ab + ac + ad + bc + bd + cd) given that a, b, c, d are distinct well (a + b + c + d)^{2} = 81 = (a^2 + b^2 + c^2 + d^2) + 2C basic rearrangement we have 3(a^2 + b^2 + c^2 + d^2) \ge 2C so 81 \ge (8/3) C or C \le 243/8 the equality case is a = b = c = d so we ignore it, giving C < 243/8 set a = 1, b = 0 and we have C = ab + ac + ad + bc + bd + cd = (c + d) + cd where c + d = -10 obviously we just let c be a huge negative number and d be a huge positive number and we don't have a lower bound so C < 243/8 unless i'm totally ****ed up right now third one edit: oh we fix the order okay so there are eight slots between them that can be +, *, or nothing for 3^8 possibilities if you want to learn how to do **** like this go to http://www.artofproblemsolving.com ps this is the best idea for a course i've ever heard where the **** do you go pps this should be a ****ing middle school or high school class not a college class ppps if you're stuck on your homework ask questions in the forum in the link pppps seriously kilga you would probably like this site it has the least retards out of basically anywhere on the internet |
Re: Creative Problem Solving
Second one can't be right.
However, I just got home from work and there's no way I'm figuring out the correct answer right now XDXD Here are two good math problems. 1) Several straight line segments are drawn on a plane surface in such a way that their intersecting lines form 1,597 areas that are not further subdivided. What is the minimum number of line segments that must be drawn to form the described pattern? 2) 1 + 10^1,234,567,890 triangles are drawn on a plane surface. What is the maximum number of areas, not further subdivided, that can be formed as these triangles intersect each other? |
Re: Creative Problem Solving
first one number of areas in which m lines divide the plane is { m \choose 2 } + { m \choose 1} + { m \choose 0 } = m(m+1)/2 + 1
gay calculation gives m = 56 exactly o tricky second one too lazy to do whatever ps hey guys itt we post problems - find all positive integers n such that 5^5 - 5^4 + 5^n is square - find the maximum and minimum value of 3 sin x + 4 cos x - find the largest even number that can't be expressed as the sum of two positive odd composite numbers - find all positive integers n such that 2^n - 7 is a square |
Re: Creative Problem Solving
Interesting method XD
I just derived .5x^2+.5x+1 = 1597 from a sequence of the first 4 lines = 56 lines, correct. Second one is tricky :D |
Re: Creative Problem Solving
number 2 is rite but there's a much easier way to do it
number 3 is rite obviously |
Re: Creative Problem Solving
sup riddles site
http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml 'putnam' is the pure maths section, 'hard' would probably also be worth checking out. |
Re: Creative Problem Solving
second problem that kilga posted would probably have been a lot easier with calculus is that what you were thinking of
hey putnam i'll be doing that in college lol |
Re: Creative Problem Solving
yeah basically
take the double derivative, set it equal to zero, find all c that makes the determinant greater than zero |
Re: Creative Problem Solving
yeah thats what i thought but screw calculus basically none of the competitions i do allow it lol
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Re: Creative Problem Solving
o I read the question incorrectly.
See, I forget almost everything I've learned in calculus like 2 months of being outside the class and not using it. XD |
Re: Creative Problem Solving
mathmania
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Re: Creative Problem Solving
AAAAing is hard
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Re: Creative Problem Solving
yeah I want a class like this too wut da fux =(
ps how many sets S=({A},{B},{C}) can you have such that A U B U C = {1,2,3,4,5,6,7,8,9,10} and A, B, and C are disjoint pps artofproblemsolving.com more like if your post doesn't solve a millenium problem you get reported for spamming |
Re: Creative Problem Solving
Here's a couple more we went over today. Two of them I figured out last week but the third got me.
1. Prove that if p is prime and (p^2)+2 is prime, then (p^3)+2 is prime and (p^4)+2 is prime. 2. If you divide a prime number by 30, is the remainder necessarily prime? 3. Without doing calculations, determine (and prove) which of 31^11 and 17^14 is greater. |
Re: Creative Problem Solving
2. If you divide a prime number by 30, is the remainder necessarily prime?
Basically the question is -> if n(30)+x is prime, then x is a prime number for x<30. If x is a multiple of 2, 3 or 5, n(30)+x is not prime since n(2*3*5)+2x/3x/5x is always a multiple of 2, 3 or 5. If x is not a multiple of 2, 3 or 5, it is a prime number. Therefore, if n(30)+x is prime, x is not a multiple of 2, 3 or 5 and x is a prime number. So yes. |
Re: Creative Problem Solving
1. TRICK QUESTION O SNAP.
p^2 + 2 is only prime for p = 3 ;) for every other prime it's divisible by 3 ;) ;) 3. (31^{11})/(17^{14}) < (34^{11})/(17^{14}) = (2^{11})/(17^3) < (2^{11})/(16^{3}) = 1/2 so 17^{14} is bigger |
Re: Creative Problem Solving
1 is correct although I'd like to see a proof of your claim to make sure you didn't just decide 3 was the only applicable prime >_>
2 is incorrect 3 is correct but there's an easier way to do it |
Re: Creative Problem Solving
1. DA PROOF
if p = 3 we have **** that works if p isn't 3 then either p \equiv 1 \bmod 3 or p \equiv 2 \bmod 3 then p^2 + 2 \equiv 0 \bmod 3 either case (in other words, 1 is the only nonzero quadratic residue mod 3) OR IF YOU DON'T LIKE THIS HARDCORE NUMBER THEORY p^2 + 2 = (p^2 - 1) + 3 = (p+1)(p-1) + 3 so for p not equal to 3 we have either p+1 or p-1 is divisible by 3 2. TRICKY TRICKY 1 ISN'T A PRIME LULZ |
Re: Creative Problem Solving
Well, I guess I made the problem harder that what it was supposed to be, since for some reason I thought that 1 was a prime... although answering 31-->1(not a prime!) seems dumb seeing as how when you ignore that 1 isn't a prime it's much harder to work it out. So I might as well make something out of it:
If you divide a prime number by N, for what values of N is the remainder necessarily prime or 1? |
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