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Fill a 2nx2n array with the numbers from 1 to
. Calculate all the sums of two adjacent numbers (sharing an edge is considered adjacent). Suppose S is the largest sum. Express min S in terms of n.Well no.
is the coefficient of degree n-1, which when multiplied with and a constant gives us the coefficient of degree n + (n-1) = 2n-1.Okay I'll have a try.
For constant polynomials we easily get P(x) = 0 and P(x) = 4.
Suppose the polynomial is written as
( should not be 0)We will expand both sides of the original equation and consider the coefficients of the two polynomials we receive, degree by degree they should be equal.
Degree 2n:
LHS:
Thus
Degree 2n-1:
LHS:
Thus
and from now we ignore this coefficient.Degree 2n - 2:
LHS:
Thus
We can then prove all coefficients except for the highest is equal to 0.
So the formula for m can be written as m = (1/4)^(n-1).
It gets really tricky because it can be of any degree. I'm still trying to figure out a catch-all strategy.
Determine all polynomials (of all degrees) satisfying
[P(2x)]^2 = 4P(x^2) for all x.
So far I have determined the following polynomials:
Undefined degree: P(x) = 0
Degree 0: P(x) = 4
Degree 1: P(x) = x
Degree 2: P(x) = (1/4)*x^2
I believe that such P(x) exists for all degrees, and it equals m*x^n for all degrees n not smaller than 1. However, I have been able to prove neither the formula (P(x)=m * x^n) nor any formula for the coefficient m.
Help! Hilfe! Hjælpe!
For anyone still interested in the problem, here is my solution:
It is obvious that a has to be of the form 30^x. Hence x.(b^c) = 30^30 = 2^30 * 3^30 * 5^30. Thus b = 2^m * 3^n * 5^p such that m, n and p are not all zeroes (since b is not 1) and mc, nc and pc are all not greater than 30.
If c = 2, we have 16 choices (from 0 to 15) for each of m, n and p for a total of 16^3 - 1 choices (x would be uniquely determined, and 1 is subtracted since m, n and p cannot all be zeroes). For c = 3 we have 11^3 - 1 choices, for c = 4 we have 8^3 - 1, for c=5 7^3 - 1 etc.
When adding all those figures up to c = 30 we will get 7041.
The thing is I'm better with discrete problems than geometry, and so I rarely gets the solution.
I have checked the source again. It seems to actually be: (BC + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2. (First term is BC not BA).
Sorry for the fault!
In triangle ABC, the median AD, altitude BH and angle bisector CE are concurrent. Prove that (BA + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2.
Please provide detailed proof.
I trust that source, so I was just asking how can I reach that answer. Right now I can't even think of a way to get any answer.
Original problem: 2015 IWYMIC team contest, problem 4:
"There is a 4×4 grid posted on the wall. Find the number of ways of placing two
identical red counters and two identical blue counters on four different squares of
the grid such that no column or row has two counters of the same color. "
The official answer is 3960. However, I could not reach this answer. Please help.
(The website of the competition is at imc-official.com, but it is seldomly updated).
Why should 6*SQRT2 = 8.5?
Also you forgot to square c:
a^2 + b^2 = c^2
6^2 + b^2 = 6^2*(SQRT2)^2
36 + b^2 = 36*2
b^2 = 36
b = 6
OK?
How many ways are there to express 30^(30^30) as a^(b^c) where a, b and c are integers greater than 1?
I have already drawn up a detailed solution that yields 7040 ways, but can you guys please confirm through computers?
Thanks!
EDIT: I miscalculated. It was 7041.
She said, "Is there any way to do this without a computer?". I couldn't answer.
I found a better solution!
2004^2 = 2002^2 + 4*2002 + 4 = (m + n + p)^2 + (m + n + p) +4 = (m^2 + n^2 + p^2) + 2(mn + np + pm) + (m+ n + p) + 4 = mnp + 4 + mn +np + pm + m + n + p + 4 = (m+2)(n+2)(p+2).
I copied it later. Thanks!
She brought me that problem.
Wow. Didn't expect it. Thanks a lot!
I'll print it out and meet her today.
Also, did you post those boards as an image? How did you do that?
I did it with a friend. She arranged the columns of 3 numbers and I rearranged them to fit rows with sum of 40.
Yes, please.
For any given board, we can rearrange the columns and rows in 3!x5! ways.
If we call a "basic" board as one that has 1) the number 1 at the top left position and 2) have the first column and the first row both in ascending order, then I have found 9 basic boards (and thus I claim the number of boards is 9x3!x5!). Here they are:
1 - 3 - 11 - 12 - 13
8 - 7 - 9 - 10 - 6
15 - 14 - 4 - 2 -5
1 - 5 - 9 - 11 - 14
8 - 12 - 13 - 3 - 4
15 - 7 - 2 - 10 - 6
1 - 3 - 8 - 13 - 15
9 - 11 - 12 - 6 - 2
14 - 10 - 4 - 5 - 7
1 - 5 - 6 - 13 - 15
11 - 10 - 4 - 8 - 7
12 - 9 - 14 - 3 - 2
1 - 3 - 11 - 12 - 13
9 - 6 - 8 - 10 - 7
14 - 15 - 5 - 2 - 4
1 - 2 - 11 - 12 - 14
10 - 7 - 8 - 9 - 6
13 - 15 - 5 - 3 - 4
1 - 2 - 9 - 13 - 15
11 - 14 - 5 - 4 - 6
12 - 8 - 10 - 7 - 3
1 - 6 - 8 - 10 - 15
9 - 11 - 3 - 12 - 5
14 - 7 - 13 - 2 - 4
1 - 5 - 9 - 11 - 14
10 - 4 - 12 - 6 - 8
13 - 15 - 3 - 7 - 2
I think if the above nine are all the possible "basic" boards, then my answer (9x3!x5! = 6480) should be right. However, a math teacher from a specialised school disagreed. Can you find another "basic" board? Or maybe my conclusion that there could be 3!x5! rearrangements for each basic board is wrong? I'm a bit unsure on this.
How many ways are there to fill a 3x5 board with integers 1-15 such that the sum of the numbers on each row is the same; and the sum of the numbers on each column is the same as well?
My current count is 9x3!x5!=6480.
Btw is there any formula for square boards (n x n) for the same question?
I have found the proof that there are only 3 solutions: (1000, 1000, 2) and its permutations.
It is very long, so I'll post it a little while later, but the key idea is: m+2, n+2 and p+2 must all be factors of 2004^2. First note that none of the three numbers m, n and p can be -1, 0 or 1 (this can be proven with discriminant). Let a=m+2, b=n+2 and c=p+2, then they are all factors of 2004^2. Consider three cases for a, b and c: all three number are positive, two are positive and one is negative, or one is positive and two are negative. We can prove that the only possible cases for a, b and c such that a+b+c=2008 are (1002,1002,4) and its permutations.
What do you mean bobbym? You found some solutions?
Something along the lines of Algebraic Arithmetic (it isn't always clear how to translate into English).
Whatever.
It's from a math book without solutions. Sad but true.
Find all triples of integers (m, n, p) such that m + n + p = 2002 and m^2 + n^2 + p^2 = mnp + 4.
Please provide detailed solution (with mathematical proof).
All of this started after I read the answer for the following question:
"Given are fifty square cardboard pieces and fifty equilateral triangular cardboard pieces, all of side length 1 cm. Construct a set of different convex polyhedra with these pieces as faces. Two polyhedra with the same numbers of vertices, edges, square faces and triangular faces are not considered different. Use as many of the 100 pieces as possible".
The answer included: the regular tetrahedron, the cube, triangular prism, J1, J7, J8, J14, J16, the cuboctahedron and the small rhombicuboctahedron.
I was curious how many polyhedra could be constructed with only square and equilateral triangular faces, and so eventually, with some research, came to the following questions:
1) How many convex polyhedra have only faces that are squares or equilateral triangles as faces? (+list)
2) How many convex polyhedra have only faces that are regular polygons not triangles? (+list)
3) (Bonus) Does the set of convex polyhedra with only regular polygon faces include only Platonic, Archimedean and Johnson solids, plus regular prisms and antiprisms?
Some additional clarification:
1) Polyhedra that only has squares or only had equilateral triangles are also included (e.g. cube, regular tetrahedron)
2) I will be discounting regular prisms (which are infinitely many in number), except the cube, which, well, is special. Anything that includes only regular polygon faces with at least 4 edges.
Thanks for your help.