Puzzles with numbers and things

[snookersfun]

update: fiendish battlechalks 405 solved by Mon and moglet, since yesterday actually. Congratulations!

R.406 1-9 ball setuplittle trinagle-1.jpg

Charlie found some pool balls in Barry's shop. He decides to use several 1-9 balls for setting up a triangle of base 9 in such a way that each possible straight line formed by touching balls contains different valued balls.

To let Oliver, Gordon and Gwenny have a go at figuring out his arrangement he then provides them with the following drawing. The numbers added in by him between triplets of balls equal the sum of the values of the three balls in each triangle's vertices.
ball grid-1.bmp

What is Charlie's setup? Answers by PM please!

[moglet]

R403

[ATTACH]2631[/ATTACH]

R404

Youngest first:

Now: 13,16,18**,19,23,43

The year Mike was born: (0),3,5,6**,10,30

The year Rob was born the coach was 20

When the coach is 50 Mike will be 20

The year J was half Marl’s age now: 6,9**,11,12,16,36

** 6x3=18, 18/2=9

[snookersfun]

nothing more to add about R.402

but the solutions to R.403 and 404 can come up on the thread as well now (non-PMers invited too)

and...

R. 405 Battle-chalks

This one is a whole lot like R.399, which now seems ages ago already.

In the following 12x12 grid position several linear formations (horizontally or vertically) of triangle chalks, as indicated on the side of the drawing, in such a way that none of the shapes touch neighbouring ones (not even on corners).
battlechalks-2.bmp
The numbers placed in the grid give the amount of cells occupied by pieces of triangle chalk
a) as the sum in respective row and column where the specific number is placed and also
b) when reading the numbers standing on head (or rotating the picture 180°) as the sum in the two lines diagonally crossing through the respective numbered spots.
No chalks can be placed on the numbered spots themselves.

Answers by PM please

[abextra]

Thank you for the explanations, Monique and Moglet!

This is how I did it.

Let's say the size of Charlie's pile was a.
Oliver's pile size then was b=(146a-1)/147
Gordon's pile size was c=(146b-1)/147=(146²a-146-147)/147²
Gwenny's pile size was d=(146c-1)/147=(146³a-146²-147*146-147²)/147³

I was stupid enough in the beginning to calculate the values of 146² and 147² etc all the way, so at this point I had an equation

3112136a - 3176523d = 64387 and was stuck for days, lol. Then replaced 147 by x and 146 by y and got

y³a - x³d = (x²+xy+y²) = x³-y³

and from that 147³/ 146³ = (a+1)/(d+1), which means Charlie's pile size was 147³-1 balls and Gwenny's pile size was 146³-1.


Not sure this makes sense for you, my dear geniuses...

[moglet]

More about R402.

Barry's shop must be more like an aircraft hangar, the number of balls necessary for the mischief to work would fill a cube shaped crate forty metres on a side (ref:R361 for densest packing actually about 37m!). Had there been six mischief makers, the crate would have been over one kilometre on a side.

I started with this algebraic transposition:

(a^n - a)/a= a^(n-1) - 1

because:

((a^n - a) - (a^(n-1) - 1) - 1*))/a= a^(n-1) - a^(n-2) - 1

If a and n are integers and a is greater than n and n is greater than 1, then the subtraction of 1* (the ball each ape keeps) and the quotient (the hidden cache in the cellar) in the first equation, then the next and successive subtractions will, I think, always be divisible by "a" without remainder "n" number of times.

All that remains is to decide on the minimum size of the original pile. We know the pile has to be of the form:

a^n - a + 1*

If we set a=147 (the number if smaller piles) and n=number of apes we get an original pile of:

147^4 - 146

If we use a lower power the last ape will have a pile that will not divide by 147 without remainder after subtracting the single ball. The cached pile sizes are given by:

1) 147^3 - 1
2) 147^3-147^2-1 or 147^2x146 - 1
3) 146^3+146^2-1 or 146^2x147 - 1
4) 146^3 - 1

Resolving the three smaller cached piles was somewhat tiresome for me due to the confusion with adding/subtracting of the single ball.

[Monique]

OK ... here about R402.

Let's n be the size of the huge pile standing in Barry's shop when the apes arrive.

If a is the size of Charlie's pile then we have:
a=(n-1)/147 or 147*a+1=n or 147*(a+1)=n+146
If b is Gordon's pile size, we have:
b=(146*a-1)/147 or 147*b+1=146*a or 147*(b+1)=146*(a+1)
Similarly if c and d are the sizes of Oliver's and Gwenny's piles we have:
147*(c+1)=146*(b+1) and 147*(d+1)=146*(c+1)

therefore 147*(d+1)=(146^3/147^3)*(n+146)
or n+146=(d+1)*(147^4/146^3)

because n, a, b, c and d are all integers and GCD(147,146)=1, d+1 must be divisible by 146^3 and to be minimal it must be exactly equal to 146^3

The rest follows.

[abextra]

Charlie, Gordon, Oliver and Gwenny just arrived in Barry's shop and there in the middle of the shop is a huuuge pile of balls.

...

What's the smaller possible size of the original big pile?
What's the size of each of the smaller piles hidden in the cellar?

Ok, here are the huge numbers I got:

In the beginning there were 466948735 balls in the middle of Barry's shop.
In the end there were piles of 3176522 balls (Charlie), 3154913 balls (Oliver), 3133451 balls (Gordon) and 3112135 balls (Gwenny) in the cellar.

Not sure how to explain, a lot of formulas it was, maybe Snookersfun or Moglet can help?
Or maybe someone else?

[snookersfun]

breaking news: R.404 solved by Mon and abextra! Congratulations!

...and moglet cracked this one too, well done!

Anybody else?

[snookersfun]

update: R.403 solved perfectly by abextra already. Very well done!!

maybe I need to clarify though that the loop cannot be intersecting itself.

and Mon and moglet followed suit now. Congratulations!! and apologies for the confusion

[snookersfun]

I suppose this (in bold) reads J's because it's Marl speaking ...

oh, right, certainly J.! Will correct it in the puzzle now.

And I don't understand how the paths are constructed in R403.. Can you give examples of "legal" and "illegal" paths, please?

and here is a smaller example of a solved puzzle similar to R. 403
loop.bmp

[Monique]

and R. 404: silly wording, silly title

Coach Green, a bit beyond his best snooker years nowadays, takes the new kids out for a training session. The young stars are Rob Beaker, Ramble Sky, Marl Ankle, J. Clincher and Mike Yellow. Bob Runner is around and decides to interview Marl Ankle, the age-wise middle one of the 5 up and coming superstars, after the session.

‘Just looking at that practice session, you certainly all potted the balls extremely well today. You seem to have a bright future ahead of you. How old are you now?’

Marl, always eager to show a bit of math skills above the usual calculation of potting angles and addition of balls (preferably to reach 147), decides not to give a straight answer.

‘Well, you know that Rob is the oldest of us youngsters and Mike the youngest. Ramble is older than me and J. is younger. I am now three times as old as Ramble was when Mike was born. Actually, strangely enough, Rob’s, Ramble’s Marl’s and my ages were all factors of Coach Green’s age when Mike was born. Thinking about that, the only other year where the ages of more than two of us were factors of Coach’s age was when J. was half as old as I am now.’

‘So? How old is Coach Green now?’

‘Oh, another funny thing, when Coach will turn 50, Mike will be the same age Coach was when Rob was born.’

Assuming that Marl was talking in whole years, how old is everybody? (excluding Bob)

Answers by PM please

I suppose this (in bold) reads J's because it's Marl speaking ...

And I don't understand how the paths are constructed in R403.. Can you give examples of "legal" and "illegal" paths, please?

[snookersfun]

and R. 404: silly wording, silly title

Coach Green, a bit beyond his best snooker years nowadays, takes the new kids out for a training session. The young stars are Rob Beaker, Ramble Sky, Marl Ankle, J. Clincher and Mike Yellow. Bob Runner is around and decides to interview Marl Ankle, the age-wise middle one of the 5 up and coming superstars, after the session.

‘Just looking at that practice session, you certainly all potted the balls extremely well today. You seem to have a bright future ahead of you. How old are you now?’

Marl, always eager to show a bit of math skills above the usual calculation of potting angles and addition of balls (preferably to reach 147), decides not to give a straight answer.

‘Well, you know that Rob is the oldest of us youngsters and Mike the youngest. Ramble is older than me and J. is younger. I am now three times as old as Ramble was when Mike was born. Actually, strangely enough, Rob’s, Ramble’s, J.'s and my ages were all factors of Coach Green’s age when Mike was born. Thinking about that, the only other year where the ages of more than two of us were factors of Coach’s age was when J. was half as old as I am now.’

‘So? How old is Coach Green now?’

‘Oh, another funny thing, when Coach will turn 50, Mike will be the same age Coach was when Rob was born.’

Assuming that Marl was talking in whole years, how old is everybody? (excluding Bob)

Answers by PM please

[snookersfun]

and on to
R. 403: placing crates

Barry the Baboon has a new shipment of balls in. He is now just back from his warehouse, where he distributed the crates (full of red, yellow or green balls- colour coded) according to the attached floor plan, apparently in some random order. But he now explains to Charlie and the rest of his ape friends that they could actually reconstruct the route he took.
crates-2.bmp
‘Look, I stayed on a single closed loop, while moving my crates (obviously passing through each place that denotes crates placed) while moving only horizontally or vertically (basically passing through the centers of adjacent squares). Whenever I reached a place to position a crate I would make a 90 degree turn and extend the path in that direction exactly by the value of ball colour (either coming from or getting to).

So, which path did Barry take?
Answers by PM please

[snookersfun]

And now Abextra has solved R402 perfectly aswell... Well done!

Solutions for R401 and R402 on the thread anyone?

will contribute the R.401 solution: after first having tried to figure out the smallest possible square made up from 147 smaller squares, but then finally getting the point:

For any amount m of sub-squares aimed for, one can choose m-1 sub-squares such that their sum is an odd number (i.e. by choosing an odd number of odd squares within them). This odd sum has to be the difference between two squares (as for any given odd number exist two squares differing by that number, as the difference between following squares is odd and increasing by 2 each time, n² = (n − 1)² + (2n − 1)). Therefore adding the lower of those 2 squares to our m-1 squares sums the m squares to yield the higher of the two squares.

Also, as Mon prodded forever more:
let the sum of the m-1 squares be of the form 2n-1 (as odd), and a and b be the 'final' pair of squares, so then we have
S+a²=b² or
S = (b-a)(b+a) now with b=a+1 and the above:
2n-1= b+a, therefore 2n-1=2a+1 or a=n-1, b= n

and to prove that a² is different for all previously chosen square numbers:
S= a² (+ possibly sum (of smaller squares)), with a² biggest square chosen,

therefore S≥a² and 2n-1≥a² and S/2 or n or n-1 all >a for all a≠ 1 or 2.

[Monique]

And now Abextra has solved R402 perfectly aswell... Well done!

Solutions for R401 and R402 on the thread anyone?