Puzzles with numbers and things

[chasmmi]

My guess wasnt even noticed .

It even looks like it may have been rightish before the new rules. :,(

[April madness]

They will need to guess the answers.

Nah, guessing the answer is easy, but guessing the proof

[snookersfun]

yes, but I was rightish before that http://www.thesnookerforum.com/showp...count_311.html

but with DG you always have to proof, what you throw about

[April madness]

but with DG you always have to proof, what you throw about

therefore I say it's the discrimination of those unburdened with any education


just kidding

[chasmmi]

is it something silly like just because 7 is always an even number plus a prime number and the prime number needs two square numbers at least to make it (eg 4 plus 1) therefore a remainder of seven requires the one square number to leave the prime plus the two more to collate the prime making three numbers.

Then you have to add the squares needed for the bit after the remainder seven which will be at least a square number meaning that this mimimum of one plus the other three equals four plus the white ball.

(I know this is wrong but i will feel clever for about 3 mintues.)

[davis_greatest]

ok here is basic not correct answer: 5

31x31 blue
6x6 green
1x1 black
1x1 red
1x1 white

Yes, chasmmi, this is one possibility showing that 5 colours is possible. The challenge is then to prove that nothing less than 5 is possible.

[davis_greatest]

My guess wasnt even noticed .

It even looks like it may have been rightish before the new rules. :,(

All guesses are noticed! It just sometimes takes me a while to slot in answering them all, while coping with diversions and other distractions called "work". And I don't always answer them in the order they appear.

[davis_greatest]

is it something silly like just because 7 is always an even number plus a prime number

7 = 6+1 .... 1 is not a prime number

Then you have to add the squares needed for the bit after the remainder seven which will be at least a square number meaning that this mimimum of one plus the other three equals four plus the white ball.

Well, the problem with that argument is that you are assuming that you have to make 7 on its own, and make separately whatever is left after deducting the 7. But there is nothing to say you have to do that. It would be a bit like claiming that you need at least 4 square numbers to add up to 100, because 100=50+50 and each 50 needs at least two squares. This would overlook the fact that we could say that 100 = 10^2 or 6^2 + 8^2, for instance.

Nice try though!

[snookersfun]

do the last diggits help here?
Like, every square has to end in 00, e1, e4, 25, o6, or e9 (e-even, o-odd).
So, only prooving meanwhile for a number ending in ...999:
we need to try to find three squares adding up to ..99,
obvious choices like o6+e4+e9 or 00+e4+25 or 00+e9+e9 add up to e9, so no good. Therefore 4 square numbers are needed.

Am not convinced myself, but go to eat first....

[snookersfun]

should be 00+00+e9 of course

but e9+e9+e1=o9???? just great

and for the possible couples in the same way then btw.:
also 00+e9 or e4+25 will give e9....

[davis_greatest]

You're on the right lines, but instead of looking at the last digit in base 10 (i.e. the remainder when dividing by 10), try looking at the remainders of squares after dividing them by 8.

Example, 6^2 = 36 leaves remainder 4 after dividing by 8.

[snookersfun]

blimey, reminders are only 1 and 4, therefore to construct a sum with remainder 7 at least four such numbers (4,1,1,1) have to be added up.

[davis_greatest]

blimey, reminders are only 1 and 4, therefore to construct a sum with remainder 7 at least four such numbers (4,1,1,1) have to be added up.

Excellent. Almost faultless (except they are remainders, rather than reminders - perhaps you were still thinking of my latest book on enhancing your memory?) - and except for the fact that the remainder could also be 0. So the possible remainders of square numbers, after dividing by 8, are 0, 1 and 4.

Any even number 2k, when squared will give 4k^2, which is therefore divisible by 4 so gives remainder 0 or 4 when divided by 8.

Any odd number 2k+1, when squared will give (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1.

Since either k or k+1 must be even, k(k+1) is even so 4k(k+1) is divisible by 8.

Hence the square of any odd number gives remainder 1 after dividing by 8.

(e.g. 9^2 = 81, which is 10 x 8 + 1. Or 17^2 = 289 = 36 x 8 + 1)

Since the remainders are 0, 4 or 1, you need to add at least four of them to get 7.

Well done... scoreboard to follow!

[davis_greatest]

New entrant at the bottom of the scoreboard

Congratulations:

Round 25, Vidas!
Round 26, snookersfun!
and most of all, Round 25½, davis_greatest! (half rounds are worth 2 points)


HERE IS THE SCOREBOARD AFTER ROUND 26

snookersfun……………………….…..13
Vidas……………………………………….7½
robert602…………………………………4
abextra……………………………..…...3½
davis_greatest…………………..……2

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[snookersfun]

I had one remainder though ... in the sentence.

Pheew, so the world is round yet again. I can go to sleep, without tormenting my poor old brain further with math (which I might never need in real life).

But seriously, very entertaining stuff (I am odd in that way) and I learned loads today about squares and mods and excel and Lagrange