Puzzles with numbers and things

A wild guess! 73?

[elvaago]

I'm going to stick to the trend and say 70!

I shall rename that last round Austrian delight in your honour!

Pity I can only use 4 smilies, as this doesn't do it justice.

[davis_greatest]

Round 95 - Austrian delight II, with more balls

Charlie has now come for a game of Chimpsnooker. He has the most balls of all - in fact, he plays Chimpsnooker with a triangle of 20 rows of reds, instead of the usual 5 rows (so that's 210 reds in the triangle, instead of 15).

But silly Gordon has set up the triangle upside down again! Does he never learn? So what is the fewest reds that need to be moved to get the triangle the right way up this time?

[davis_greatest]

Congratulations, austrian_girl!

My bid for question a would be 5. For b, 18. Don't think it's the lowest possible but maybe proves a challenge to some.

That is the correct answer! (I mean that 5 and 18 are the correct answer - not that the statement that they are not the lowest possible is the correct answer, because they are the lowest possible!)

I shall rename that last round Austrian delight in your honour!

SO HERE IS THE SCOREBOARD AFTER ROUND 94

snookersfun.........................47
abextra...............................31
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
austrian_girl and her dad.........3½
Semih_Sayginer.....................2½
Snooker Rocks! .....................2½
Ginger_Freak.........................1½
April Madness........................1

ROUND 95 ...

... to follow.

[Snooker Rocks!]

I can get 5 for the first....and working on the second.

[elvaago]

And the 18 on the 10 row one. Impressed by 15.

[elvaago]

I got the 5 on the 5 row one too. :-)

[Robert602]

Think I can do 5 and 15

Edit: Forget that, I've drawn the wrong size triangle Now I get 5 and 18 too.

[abextra]

I think, austrian_girl is right. I haven't found anything lower either.

[elvaago]

I can't do any better than 6 and 21.

My bid for question a would be 5. For b, 18. Don't think it's the lowest possible but maybe proves a challenge to some.

[davis_greatest]

Round 94 - Australian delight

I have noticed that recent Australian birthday boy Sarmu hasn't answered a question for a few rounds, so here is one he should be good at. It's about turning things upside down.

Round 94 - Australian delight

Gordon is playing snooker with Oliver. However, as Oliver is about to break, he notices that Gordon has set up the triangle of reds upside down - i.e. the single apex red is nearest to the black and the row of 5 reds is nearest to the pink!

Question (a) What is the smallest number of reds that Oliver needs to move to make the triangle the right way up? (As long as the triangle is the right way up, it does not matter how close to the pink or black it is.)

Question (b) Later that evening, after a few banana cocktails, they are playing again but this time it is a variation of snooker - instead of 5 rows of reds in the triangle, there are now 10 rows! (So, 55 reds instead of 15.) Again, Gordon once accidentally puts the triangle upside down. How many reds does Oliver need to move, this time, to make the triangle of 55 reds the right way up?


Answers can be posted on this thread. Hopefully, there will be a correct answer by the time I return from a marathon snooker session myself, in about 16 hours from now.

[davis_greatest]

Round 93 solution

I shall now close round 93.

The answer is 2 presents. Charlie can easily seat people to prevent me from matching 3 presents with their intended recipients (for example, by making everyone sit in the same order as the presents, but going round the other way, so clockwise rather than anticlockwise). However, no matter where everyone sits, I shall always be able to match at least 2 presents by rotating the table.

Here's why:

At first, not one of the 16 guests is matched to a present. I can therefore rotate the table through 15 other positions, and as it passes through all those positions, everyone must be matched up once with a present on the way. That's 15 positions, and 16 guests matched to a present - so at least one position must match at least 2 presents to their recipients!

I shall award a whole point to snookersfun, whose (final!) answer was good enough (although not as elegant as the above ); and half a point to each of abextra and Snooker Rocks!, who both said the answer was 2 but did not satisfactorily prove it.

SO HERE IS THE SCOREBOARD AFTER ROUND 93

snookersfun.........................47
abextra...............................31
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2½
Ginger_Freak.........................1½
April Madness........................1

ROUND 94 ...

... to follow.

[davis_greatest]

Round 93... update and hints

Well, I've had a variety of answers sent to me - some better than others, but none wholly satisfactory yet

Some are saying that

(a) Charlie should be able to find a way to make people sit so that I can only line up one present with its intended recipient;

while some are saying that

(b) I should be able to match two presents to their intended recipients.

Well... I like the answers (b) more BUT the only answers I've had showing two recipients have given a single specific way of arranging the partygoers at the table - with it then possible to match 2 presents.

For those answering (b) (which is correct ), you need to show that Charlie could not possibly find any arrangement that would prevent me from matching more than one present. It is not good enough for you simply to state one particular arrangement that allows me to match 2 presents - that does not prove that there is no other arrangement for which I could not match 2 presents. (Otherwise, you might just as well seat everyone to the right of his/her present and say that the answer to the question is 16.)