Puzzles with numbers and things

[snookersfun]

Where you say one digit between the ones, two digits between twos etc, do you mean digits or numbers? What happens if we get up to 10 or more so the numbers are 2 digits long?

I'll start with an opening bid of 37.









Just kidding I've only got 4, so far, which is the same as in the question!

right, it should be numbers, will correct that now (had that written originally and though digits sounds better, but it is wrong!... )

So, 4 meanwhile the best bid, as long as it is different from mine, it'll do meanwhile...

[davis_greatest]

right, it should be numbers, will correct that now (had that written originally and though digits sounds better, but it is wrong!... )

So, 4 meanwhile the best bit, as long as it is different from mine, it'll do meanwhile...

Actually, my one for 4 happened to be the same as yours - but I can always reflect it from left to right if you like!

But I'll get something higher when I get back from shopping tonight

[snookersfun]

Hmm. I've been working at this for a while now

I wonder if there is one possible sequence for any set of numbers

there are possible sequences apart from and higher than n=4, so keep trying

[Snooker Rocks!]

For every integer of n?

So if n=9 it would work?

[davis_greatest]

For every integer of n?

So if n=9 it would work?

Not every integer. Not sure about 9.

By the way, I am extending the Initial Deadline for round 86 by 24 hours. The Final Deadline will be 23:00 GMT tomorrow, 22 December. We've had one answer.

There will be no further extension to the deadline of round 87. We've had 6 correct answers so far. About 6 hours to go till the Deadline of 23:00 GMT today, for anyone else seeking a point!

[davis_greatest]

Round 87 has closed!

Congratulations to Ginger_Freak, abextra, Snooker Rocks!, snookersfun, chasmmi and elvaago for correctly identifying the answer to round 87 to be the self-descriptive number 6210001000 - this is a number with 6 zeroes, 2 ones, 1 two, no threes, fours or fives, 1 six, and no sevens, eights or nines.

You are reminded that, unlike the other quizzes on this forum, if two or more people have the same number of points, the person who scored first will be higher (or highest) placed.

Since April Madness was already on one point, she is placed above new entrants Ginger_Freak and Snooker Rocks!. Since Ginger_Freak submitted an answer before Snooker Rocks!, Ginger_Freak is placed higher than Snooker Rocks!.

SO HERE IS THE SCOREBOARD AFTER ROUND 87 BUT BEFORE ROUNDS 86 AND 88

snookersfun………..………….…..43½
abextra..............................25
davis_greatest....................19
Vidas.................................12½
chasmmi.............................10
elvaago...............................9
Sarmu.................................8
robert602.............................6
The Statman…………………. …...…5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
April Madness........................1
Ginger_Freak.........................1
Snooker_Rocks! ....................1

[davis_greatest]

I’ll put up another question, round 88:
You should try to arrange the numbers 1,2,3,….n (two of each) in such a way, that there is one other number between two ones, two other numbers between two 2s, three other numbers between two threes, etc.

Example: for n=4 i.e. using 1,2,3,4 twice each: 4,1,3,1,2,4,3,2
Try to get as high a sequence as possible in this way. Bids for number of n reached (so ½ the digits in the sequence) can be put up on the thread until Sunday 24.12. 12:00pm GMT. After that I’ll ask for the actual sequences.

Will I spoil this round if I make an opening bid of 11?

[davis_greatest]

Round 88 - my next bid is 20

[davis_greatest]

OK - my next bid to snookersfun's round 88 is 24, but keep bidding even if you bid lower than this.

If no one beats my bids, I shall ask snookersfun to award a point also to the person with the next highest bid.

Edit: I'm changing my bid again. I now bid 40.

[snookersfun]

OK - my next bid to snookersfun's round 88 is 24, but keep bidding even if you bid lower than this.

If no one beats my bids, I shall ask snookersfun to award a point also to the person with the next highest bid.

Edit: I'm changing my bid again. I now bid 40.

...and there was I, being afraid of never getting any bidders...
impressive!

but as d_g said, keep bidding! So far, any other bid above 4 would be eligible for a point, or maybe half, depending on the final outcome....

[davis_greatest]

I think I'll up my bid to 76

Edit: 84

[davis_greatest]

Round 87 explanation

Here, by popular request, is a brief explanation of how you can get the 6210001000 answer for round 87. It's not a full solution - more of an outline - but hopefully should help.

The problem can be solved by first trying 9 zeroes, then 8, then 7, then 6.... you will find it words when we get to 6 zeroes.

First, note that the digits must add up to 10. That's because there are 10 digits, and each digit corresponds to the number of times that a digit appears!

Similarly, any digit multiplied by the 'digit that it represents' cannot be more than 10. For example, there cannot be 2 or more 9s, because 2x9 = 18 is more than 10, and the digits must add up to 10.

That means the last four digits can only be 0 or 1.

So, for instance, the final digit (representing 9s) could only be 0 or 1. If it were 1, that would mean there must be a 9, which means a digit must appear 9 times. This would have to be the first digit (representing zeroes). So the number would look 9000000001. But that has 8 zeroes, not 9... and it also would need a 1 in 2nd place.

So the last digit must be 0. If you try a 1 in the last-but-one place (representing the 8s), there must be one 8, which means 8 zeroes. So you would get 8000000010 (which was what elvaago offered first! ) But this doesn't work because there is a 1, and so the 2nd digit must be a 1.

So the number ends 00
Try a 1 in the place representing 7s.... you get a similar problem.

So try a 1 in the place representing 6s, so the number ends 1000

So we will have 6 zeroes, so the number is 6xxxxx1000

The first x can't be zero, because there is already at least one 1. It can't be 1 either, because if we made it 1, there would be 2 ones, whch would mean it would need to be a 2!

So we can write 62xxxx1000

There is now one 2, so the 3rd digit must be 1, and we find that 6210001000 works perfectly. The only solution!

Congratulations to Ginger_Freak, abextra, Snooker Rocks!, snookersfun, chasmmi and elvaago for correctly identifying the answer to round 87 to be the self-descriptive number 6210001000 - this is a number with 6 zeroes, 2 ones, 1 two, no threes, fours or fives, 1 six, and no sevens, eights or nines.

[Snooker Rocks!]

D_G_E has obviously got a method to solve this (round 88)!!

[davis_greatest]

D_G_E has obviously got a method to solve this (round 88)!!

Hehehe - yes! A computer! I wrote a short program which can come up with sequences as long as I want They can be done by hand too, but that takes rather longer. I'll send it to anyone who wants it.... once bidding has closed.

[davis_greatest]

and I'll give a quick hint as snookersfun isn't here... don't waste your time trying to find a sequence with 5 pairs or 6 pairs... those are impossible.