Puzzles with numbers and things

[davis_greatest]

I could join the dots with 8 straight lines...

8 is good... in fact, I haven't done it with any fewer! Let's give until Monday 10a.m. GMT for anyone to match or beat 8 lines by pasting a picture on the thread.

PS Your dots are neat enough IMO.

Thank you, abextra!

While round 279 More dots? remains open, here is...

Round 280 Square balls

Charlie goes to the warehouse at the back of Barry The Baboon's Ball Shop, carrying a sack of bananas, and sees a nice arrangement of balls that Barry has put on display. The balls are arranged in a solid Big Square, which is divided into four equal Small Squares of balls: one brown, one blue, one pink and one black.

"That's nice," says Charlie. "When did you get all those balls?"

"This morning," replied Barry. "The balls arrived in big buckets, each containing an equal number of balls banging together. Once all the buckets had been delivered, I discovered that I could make this nice arrangement, and had exactly one ball left over!"

"How many buckets were there?" Charlie asked.

"I don't remember," replied Barry, "but I do remember noticing that the number of buckets was more than the number of balls in each bucket."

Charlie examined the balls more closely, and noticed that the number of rows of balls in each Small Square was equal to the square of the number of bananas in his sack!

"In that case," said Charlie, "the number of buckets must have been exactly 100 more than the number of balls in each bucket!"

How many bananas in Charlie's sack?

Answers initially by Private Message please...

[abextra]

Round 279 - More dots?

Same question then... but 25 dots in a square. Join them up with as few lines as possible, without removing pen from paper.

PS If anyone knows how to do neater dots than mine, feel free to put up a picture.

I could join the dots with 8 straight lines...

PS Your dots are neat enough IMO.

[davis_greatest]

Congratulations Monique and snookersfun!

So....

Round 279 - More dots?

Same question then... but 25 dots in a square. Join them up with as few lines as possible, without removing pen from paper.

PS If anyone knows how to do neater dots than mine, feel free to put up a picture.

[snookersfun]

here's a pic, needing running 'over' only once:
HappyBirthday41.gif

[Semih_Sayginer]

that seems spot on Monique

[Monique]

No picture ... but an explanation.

If your points are numbered

13 14 15 16
09 10 11 12
05 06 07 08
01 02 03 04

trace a line going through 1,2,3,4 and continue over 4 until
you can trace a diagonal throuh 8,11,14; again you continue over 14 until
you can trace the vertical though 13,9,5;
at 5 you trace another diagonal 5,10,15 continuing over 15 until
you can trace the vertical throuh 16,12,8
at 8 you go for an horizontal 8,7,6.

[davis_greatest]

Yes, looks impossible in fewer than 6 lines. Someone please put a picture with 6 lines.

[snookersfun]

think we all have 6

[Semih_Sayginer]

Anyone want to match (or even beat) it?

i want to, but i cant.

[davis_greatest]

I have a bid with six lines.

Anyone want to match (or even beat) it?

[Monique]

I have a bid with six lines.

[davis_greatest]

Round 278 - Gone Dotty

Here is a slight extension of a well-known puzzle.

Join these 16 dots up by drawing as few straight lines as possible. You must draw the lines without taking your pen off the paper / screen (i.e. they must be joined up).


(PS I haven't thought what the answer might be yet, but I'm sure someone else will find it!)

[Monique]

Just to add yet another view ...
Here was my reasoning after having read hint 1.

If N is any integer positive number that can be expressed as a sum of consecutive positive integers then

N= n + (n+1) + (n+2) + ... + (n+k) for some n,k >=0
or N= (k+1)*n + (k+1)*k/2

or 2*N=(k+1)*(2*n+k)

one of the two expressions (k+1) or (2*n+k) is odd, the other is even.

If r is an odd factor of N and you equate r to the odd expression ((k+1) or (2*n+k)), the other one is determined and you get a system ... to solve to find k and n ...

if N=9! it has 20 odd factors: 1,5,7,35 and their "combination" with 3,9,27 and 81.

[snookersfun]

- the break 89 to 856 corresponds to the odd divisor s=3³ x 5 x 7 = 945, and q=472. Here, t=384 (since 362880 / 945 = 384), so the break begins at q-t+1 = 89.

- the break 1290 to 1545 corresponds to the odd divisor s=3^4 x 5 x 7 = 2835, and q=1417. Here, t=128, so the break begins at q-t+1 = 1290.



That wouldn't make any difference to the number of breaks, as there would would still be the same number (20) of odd factors. So, if n=s.t, every odd factor s would be the same as before and every t would be 4x bigger than before. There would still be 20 breaks, but they would be different from before.

ahhhh, right, now I see, slight mistake in the 'even divisors' there. Those are lost on multiplying by 4 etc...(mentioned I hadn't thought through the end)
In which case all the differences will be odd finally.

[PaulTheSoave]

SF is still full of adrenaline after last night! Just ignore this thread for a couple of hours.

ehm... wot does that all mean?