1. Originally Posted by grapes
Garrggg, how'd I miss this one? Take the hexagon and place one point at the center. Then the ratio is the square root of three, 1.732...
No, you fool, the longest distance remains 2. Ouch

ETA: In my mind, I moved the farthest points of the pentagon along the circumference of the circle, to bring them closer together. Of course, that moves them farther away from others.

2. Puzzle #140 here.

The proof shows that it must be at least the square root of three, but the author is uncertain about whether this bound can be achieved.

I now have a proof that it can't.

I suspect that the pentagon-plus-centre solution might be the optimum, but I have no proof yet.

3. Originally Posted by Coelacanth
I'm trying to see if I can come up with a proof that your example is optimal, or a counter-example.
Take the convex hull of the six points. If one point, A, is interior to that hull, then I think I can show that of the ten triangles that it makes with the other five points, one of them must have an angle at A that is greater than or equal to 144 degrees. Total equality happens in my example of the pentagon, with A at its center.

Since , the cosine formula:

implies that

and, if , the ratio a/b is

If all the points are vertices of the convex hull, I think I can show that the longest distance between opposite pairs of points is more than twice the shortest side.

4. OK, the first one is wrong, but fixable, I think.

5. I don't know if you looked at my source, but the convex hull/minimum angle approach is what was used to establish the lower bound of sqrt(3), which occurs with a minimum angle of 120 degrees. But I can show without too much trouble (by a totally different approach) that the square root of three is not attainable.

6. Join Date
Dec 2011
Posts
11

7. Yes, it is now your turn to come up with a question.

8. Perhaps the next question could relate to the spelling of "Amateur"

BTW, grapes, even though we haven't exactly proved the question as asked, I think you can take honours for this one . . .

9. This is an easy one, easily googled if you want.

Everyone knows what a perfect number is-the sum of its divisors is equal to it. The known perfect numbers are all even, which is also well known. For a deficient number, the sum of its divisors is less than the number, and for an abundant, the sum of its divisors is greater. What is the first odd abundant number?

10. I'm glad you think it's easy

So, I have worked out the following. Call the number , and let it have prime factorisation

It is important that the be both prime and distinct.

Then the sum of all of the factors of (excluding itself) is

Since what we need is , we need

or

Each factor in the product is an increasing function of , but approaches the limit of , which is larger for smaller . For , it is , for , it is , and so on.

First result: every odd abundant number has at least three distinct prime factors.

If there are only two, then the product has to be less than , which is - not big enough. With three distinct prime factors, the product is bounded by , which is - big enough. So there may be odd abundant numbers with only three distinct prime factors.

A little quick search turns up that is abundant. This is the smallest number with three distinct prime factors, and five prime factors (that is, counting the repeats). We need three distinct prime factors, so any other solution can have at most four prime factors. It can't have three, because doesn't work, and making the prime factors bigger makes you more deficient, not more abundant. So the solution is either , or a number with four prime factors, at least three of them being distinct. If all four are distinct, the smallest possible number is 1155, which is too big, and also not abundant. So the solution is one of , , , or . The last three are all less than , but none of them are abundant.

So I'm getting . Am I right?
Last edited by Sarawak; 2012-Feb-07 at 06:37 AM. Reason: Careless math errors, not affecting final result though

11. 945 is correct. Your turn.

http://en.m.wikipedia.org/wiki/Abundant_number

Interesting stuff at that wiki, e.g., every number greater than 20161 is the sum of two abundant numbers; and the smallest odd one not divisible by 3 has ten digits.

12. Let me think about it. It's always hard to come up with a good one . . .

13. Find a solution to with , , and all prime, or prove that no such solution exists.

14. Does (3, -2, 1) count?

15. No

16. OK! Since that is the only possibility with even primes, all three must be odd primes. Which means you get a remainder of 1 when you divide the x2 term or the z4 term by 4. You can get a 1 or a 3 when you divide the y3 term by 4, but in neither case would the sum of the remainders on the left be equal to the remainder on the right. GICBD.

17. Originally Posted by grapes
OK! Since that is the only possibility with even primes, all three must be odd primes. Which means you get a remainder of 1 when you divide the x2 term or the z4 term by 4. You can get a 1 or a 3 when you divide the y3 term by 4, but in neither case would the sum of the remainders on the left be equal to the remainder on the right. GICBD.
What is "GICBD"?

Can you elaborate on how you know it is impossible with even primes?

The approach I have been taking (not finished yet) is, odd+odd=odd is impossible. Also, odd+odd=16 is impossible, since y^3 is too big even for y=3. For x^2+8=z^4, an odd prime squared must end in 1, 5, or 9, so z^4 must end in 3, 7, or 9, which is impossible - it must end in 1 or 5.

So the only possibility then is 4+y^3=z^4, and I have been working on that one

18. Oh, I forgot to exclude (x=2, y=2, z=odd), (x=2, y=odd, z=2), (x=odd, y=2, z=2), and (x=2, y=2, z=2), but they're all rather easy.

19. I think using remainders by division by three takes care of that, no?

GICBD is Guess It Can't/Can Be Done, like QED but less Latin

ETA: but why primes? Why not integers in general?

20. Originally Posted by grapes
I think using remainders by division by three takes care of that, no?
You are referring to the missing part of my solution, proving 4+y^3=z^4 is impossible for odd y, z? If so, can you elaborate? (Not quite following.)

Originally Posted by grapes
ETA: but why primes? Why not integers in general?
I copied it from a web site, which didn't mention that. There are integer solutions, e.g., x=28, y=8, and z=6.

21. Oh, I got it. I was being silly, I flipped a sign.

22. This is kind of an odd problem. My first post is obviously too much, but why restrict it to primes? Did they even have that solution, or did you find it?

One of them must be even, as you say, because the remainders wouldn't add up. For the same reason, one of them must be divisible by 3. Checking the possibilities for 2 and 3 show it can't be done. Using only primes.

23. Originally Posted by grapes
This is kind of an odd problem. My first post is obviously too much, but why restrict it to primes? Did they even have that solution, or did you find it?
My source had no solution of any kind. I went out on a limb a bit asking it, because I did not know the solution myself at the time

I think the restriction to primes is to restrict to odd numbers or two, since there are solutions for even numbers other than two.

Originally Posted by grapes
One of them must be even, as you say, because the remainders wouldn't add up. For the same reason, one of them must be divisible by 3. Checking the possibilities for 2 and 3 show it can't be done. Using only primes.
Ah, I like this one.

Remainders by two, if all numbers odd:

{1}+{1}={1} <== impossible, so at least one number even

Remainders by three, if all numbers not divisible by three

{1}+{1,2}={1} <== impossible, so at least one number divisible by three

Of course, this method of proof does not preclude non-prime solutions, even odd ones

24. Just not all odd, right?

And, y3 = (z2 + x )(z2 - x )

25. Originally Posted by grapes
Just not all odd, right?
Yes, but not all odd can be ruled out by odd+odd!=even. The possibility of including two makes it a bit more difficult.

Originally Posted by grapes
And, y3 = (z2 + x )(z2 - x )
Yes, but what if z2-x=1?

26. What is the smallest number not ending in zero that is the product of three distinct primes?

27. Originally Posted by Coelacanth
What is the smallest number not ending in zero that is the product of three distinct primes?
42

28. Correct! You're up.

29. The following integral was derived from calculations of gravity due to a spherical shell of radius 1 centered on the origin, on a point at R.

For R > 1, the integral evaluates to 2/R2; for R < 1, it is zero. What exactly is going on at R=1? (I think I know.)

30. That's a rather open-ended question

Do you want the value, or some description of the behaviour of the integrand that causes the discontinuity in R?

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