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Can you type Factorization(1111111111111111111); into Magma ? And it shows all the digit,s ?
If I type Factorization(1111111111111111111); into Magma the output is [ <1111111111111111111, 1> ]. You'll have to tell me if this is showing all the digits.
A result related to this discussion is the following:
Proof
Certainly a should be even, since if a is odd then
Suppose n = st where s is odd and s>1. Then
It follows from this result that
is divisible by so 1001 is divisible by 11.Of course, ganesh's example shows that the converse of this result is not true since
is not prime even though 10 is even and 4 is a power of 2.I think bobbym's point is that, even for computers, the calculation of large factorials is not a very good way of testing if numbers are prime.
At university, if I type Factorization(1111111111111111111); into Magma I can't even tell that there is a delay before it prints [ <1111111111111111111, 1> ] confirming that 1111111111111111111 is indeed a prime.
If I type Factorial(1111111111111111111); instead it refuses to do the calculation because 1111111111111111111 is too large.
If
, is not always odd.For example, let x = 4.
If b is an odd number and both a and b are integers greater than ten but less than 99, find an expression for x using the variables a and b to make the statements above true.
Suppose I write
I really can't see how I have failed to find an expression for x using the variables a and b to make the given statements true.
The only problem I can see with this solution is that x is not an integer when a is even, but no solution is possible in that case.
I think that there may need to be many separate solutions to find the all the possible solutions for this problem
As I admitted in my first reply, the solutions I have given are not the only solutions, so I completely agree with you that there is more work to be done in order to find all possible solutions. However, until we can agree that I have provided some solutions to this problem I have no interest in looking for others.
The problem is to find an expression for x in terms of a and b such that a certain relationship between these three quantities holds. The value of x is not uniquely determined by a and b, and in my solution I only try to find some value of x that works. It is always possible to choose x so that x+1 is a prime power, and I choose to do so.
As an example, suppose a=3 and b=1. For any prime p,
and . Hence we can choose x to be p-1 for and prime p.I think the requirement in the second part should be that a is odd rather than b.
You have probably already shown in the first part that
In general
will depend on the prime factorisation of .If the first sentence means "There are 100 gadgets. 10 are defective." then the answer is
I'm not following you. When x=3, f(3) = 121 a square.
This exactly illustrates Kurre's point. TheDude's post presents a proof that the polynomial
1. The argument has nothing to do with this. The sine function has three zeros inside the circle C(0,6).
2. I think that f is a hypothetical entire function such that
I assume that by
you mean the chromatic number of which I would normally denote .Suppose we have coloured H with d colours and we want to extend this to a colouring of G. Since G cannot be coloured with d colours we need to use a new colour for the vertex x. What does this tell you about the colours of the neighbours of x?
It's impossible! To get to the goal,you have to choose "yes" to answer "Is the other marker in a box whose text refers to cows?".
There is NO such box!
I haven't solved this puzzle myself but I can see too possibilities.
1. If the other marker is also in box 50 then it is in a box whose text refers to cows.
2. If the rule change from box 60 is in effect then you always leave by the "yes" path.
By definition, a harmonic function is a function whose Laplacian is 0.
Your proof that the sum of two harmonic functions is harmonic is fine, and is easily adapted to prove that the difference of two harmonic functions is harmonic.
Consider the function u(x,y) = x
u is harmonic but u^2 is not so the product of harmonic functions is not necessarily harmonic.
I'm sure you can find an example to show that the quotient of harmonic functions need not be harmonic.
Show that if
Then, using that the group has odd order, it's not too hard to show that
is the identity.But I did forget to mention leading zeros are allowed
It doesn't matter. The clues define a linear system with eight equations and eight unknowns that has a unique solution. You don't need to know that the answers fit together in the grid, or that they are integers.
This is a very unusual cross-number puzzle.
I think you're right that starting and finishing aren't far apart.
I would make the small complaint that the identity does not have one of the given forms.
In general, the problem is to find the partitions of n with an even number of even parts.
Your question does make sense and you are right.
To prove that
are conjugate you must find such that .You already have an element
such that .Suppose
. If you could find an element that commutes with then you might consider the element .If you are interested in a similar problem, you might like to think about the number of conjugacy classes of 5-cycles in
.I'll have to keep thinking about 7.
Jane, for any function
Here's a proof that works:
Suppose
is continuous bijection.If a < b then consider the set
.Similarly if b < a.
PS: I feel pretty silly now. My proof is really just an application of the Intermediate Value Theorem.
The proof for question 1 that has been given is almost certainly the standard proof. However here is the idea of an another proof that I think is much prettier.
Suppose S is a non-empty set that is bounded below. Let L be the set of lower bounds for S. Show that L has a supremum and that infS = supL.
When the functions are polynomials, the most important thing to know is
In your first example we have
If we want to show that
To get the answer that you have we need to observe that for x>9
I prefer the first method, since it does not require one to think about the exact numbers involved.
I cannot give working for the Big Theta example since I do not believe that it is correct.