Puzzles with numbers and things

[Monique]

Is that 81 allowed? the way I understand "numbers" not "digits" i supposed not ... D_G?

[abextra]

Needing one more "1" to get the target - 5*5*81-6-9=2010

[Monique]

let's launch the race ... 5*5*8*(9+1) + 6 = 2006

[davis_greatest]

Happy new year to all on TSF.

Next answer (or request for hints if needed, although the clue is in the question :snooker to round 352 on the thread please.

Round 353 New Year Number Magic

Using as many as you like of the numbers 1,5,5,6,8,9 (but not using any more than once), and any combination of addition, subtraction, multiplication and/or division, get as close as you can to today's target, which is 2009.

Again, answers please directly on the thread. Quickest / closest wins!

[davis_greatest]

Breaking News

First congratulations for solving round 352 go to snookersfun - well done

Edit - and Monique now joining suit!

[davis_greatest]

Round 352 - Take Five

OK, a quick one from here... :snooker: ....complete the five missing numbers in this sequence...

18916, ?????, 53233, ?????, 84956, 57646, 57273, ?????, 88996, ?????, 10511, 21131, 20122, 12512, 91341, ?????

[snookersfun]

more solutions in now:

moglet and d_g on the distances and abextra on tilting. Congratulations everybody.

goes 'sheltering' again now...

[snookersfun]

Well... are the imaginary lines between balls always horizontal or vertical? I was hoping so, but...

see, erm, no, another red herring that grid But they are straight lines.

[abextra]

R. 351 break off- distances

This time...

...you could organise all balls into pairs, so that each pair has discrete and distinct distances between them. Starting from a pair of touching balls, continuing with the next pairs exactly one, two, etc. balls apart. In addition no ball is positioned with his center on the imaginary line between a given couple.

Can you assign the pairs?

Well... are the imaginary lines between balls always horizontal or vertical? I was hoping so, but...

[snookersfun]

Thank you, Moglet! The explaining is always the hardest part for me.

yes, thank you indeed, moglet! Very nice and complete explanation (not my strong point either)

meanwhile I have a first entrant for R.351. tallguy solved it and also learnt to attach pictures in PMs. Congrats

[abextra]

A breakdown of R349, Gwennie would have gone straight to the last paragraph!

If we take the four scores as a,b,c and d and do the algebraic calculations...

Thank you, Moglet! The explaining is always the hardest part for me.

[moglet]

A breakdown of R349, Gwennie would have gone straight to the last paragraph!

If we take the four scores as a,b,c and d and do the algebraic calculations as instructed (a full stop denotes multiplication):

(a.b.a.c.a.d).(b.a.b.c.b.d).(c.a.c.b.c.d).(d.a.d.b .d.c)

collect terms:

a^6.b^6.c^6.d^6

then divide by the product of the four scores:

a^6.b^6.c^6.d^6/a.b.c.d = a^5.b^5.c^5.d^5

No remainder, red herring, as was the number and distribution of the four scores, we just needed to know they were positive and greater than 1.
We are asked for the remainder after subtracting the four score product and division by 10, anything to the left of the decimal point is not relevant.
If we let:

n = a.b.c.d

Then we have;

(n^5-n)/10

It is a property of the fifth powers that the final digit of n (a whole number) is always the final digit of its fifth power, so, there will never be any “remainder” after division by ten.

[snookersfun]

Mon solved the tilted balance as well, even with a slightly banged up head Congrats!

and with that to:

R. 351 break off- distances

This time around the apes play a proper game of snooker, or almost, as they are experimenting with different numbers of lines in the original triangle of reds. In one of those efforts involving seven lines of reds, Gordon smashes the pack (Quinten Hann like), potting two reds right away in the process, and leaving the other reds in the following positions on the table.
26 balls-1.bmp
Charlie, being his regular clever self, takes a glance and exclaims: Wait, before you play on, look how neatly you scattered those balls. As I see it, you could organise all balls into pairs, so that each pair has discrete and distinct distances between them. Starting from a pair of touching balls, continuing with the next pairs exactly one, two, etc. balls apart. In addition no ball is positioned with his center on the imaginary line between a given couple.

Can you assign the pairs?

[snookersfun]

Hi, Snookersfun!

What does Charlie mean by " discard any fractional portion"? Should I try to spell RED HERRING?

Hi abextra, It means you ignore the remainder. For eg. 7/2 = 3 if you discard any fractional portion.

The spelling of RED HERRING is correct.

and despite the red herring R. 349 now also solved by abextra and d_g.
That means we can pretty much close that round. Please somebody put up his/her solution!

[tallguy]

Hi, Snookersfun!

What does Charlie mean by " discard any fractional portion"? Should I try to spell RED HERRING?

Hi abextra, It means you ignore the remainder. For eg. 7/2 = 3 if you discard any fractional portion.

The spelling of RED HERRING is correct.