Puzzles with numbers and things

[davis_greatest]

Wow, Talk about a brain teaser!

Fortunately Excel was quite useful here for producing the break and then reversing it for me - I don't think I'd have had the patience to type that out myself!

[Lee Vilenski]

Wow, Talk about a brain teaser!

[snookersfun]

here is the number with proper spaces inserted:

741 041 431 921 521 221 021 311 211 501 401 79 69 98 88 18 08 37 27 56 46 75 65 94 84 14 04 33 23 52 42 71 61 9 8 1

now reading that from right to left (reversed numbers and all) (! and I have to admit I initially got the result, without doing that), it will be the sequence of a maximum break in snooker

[davis_greatest]

OK, Monique and snookersfun have both got the answer. Let's hear it please....

[davis_greatest]

add an 1 in the end...! ) possibly followed by 9080

Very close. As I think Monique is too (who sent me an almost-answer by PM). There are no more zeroes....

[snookersfun]

add an 1 in the end...! ) possibly followed by 9080

[davis_greatest]

Round 160 - Thorny problem

Continue:

74104143192152122102131121150140179699888180837275 64675659484140433235242716...

(Edit: when I view this on my screen, it appears to show a space between the 5 and 6 - there is no space in the number I entered, so ignore it if you see it too - I don't know why it is there.)

[snookersfun]

lol and thanks Monique. 'absentmindly' quite spot on, though one can never fault the logic of his puzzles (and trust me, I doubted a few questions initially...)
I have to admit, on first glance I only noticed that 'once each' feature (drew geometrical patterns, figured words (but hey 1 and 0 are only blanks...)), had to step back and purge brain (oh, for two hours it seems) and suddenly it was there!

[Monique]

Congratulations Snookerfun!
I must confess that the only idea that crossed my mind when reading the terms of R159 was "Here is D_G punching the keys absentmindly (but still just once each!) while trying to remember what the number of the "puzzles with numbers and things" helpline could be

Enjoy a banana split

[davis_greatest]

You typed the numbers according to their positions in the alphabet! Why, beats me though

Indeed I did - in alphabetical order: Eight, Five, Four....

Congratulations snookersfun!

yes indeed, more than parts of bananas, I hope

What? Working towards winning a banana wasn't enough?

[snookersfun]

I was just looking at the phone on my desk, trying to think of a question, and tapped in the number 8549176320.

Why did I type this number?

You typed the numbers according to their positions in the alphabet! Why, beats me though

Answer please on this thread! First correct answer wins!


(Wins what?)

yes indeed, more than parts of bananas, I hope

[davis_greatest]

Round 159 - On the phone

I was just looking at the phone on my desk, trying to think of a question, and tapped in the number 8549176320.

Why did I type this number?

Answer please on this thread! First correct answer wins!



(Wins what?)

[snookersfun]

Happy to announce that Monique has solved the second cross-number now. Well done!
I'll wait for at least one more answer though, before closing the round....

Round 154: cross-number 2

This is a standard cross-number puzzle in which all of the index numbers in the diagram have been erased, and the 22 clues have been randomized! Each clue refers to one of the 22 numbers in the diagram having more than one digit, and no two of them are equal. Furthermore, whenever a clue refers to another number or numbers, it means that those numbers are also among the 22 diagram numbers, different from the clue number and from one another. For example, "(8) Product of a triangle and a palindrome." means that the "Product" is the number of clue (8) and the "triangle" and "palindrome" are the numbers of two other clues, and all three of these numbers are different. However the numbers referred to in different clues may or may not be the same, only the clue numbers are sure to be different. Find the unique solution.

Note: All numbers are positive integers, base 10, having more than one digit. No number may begin with a 0. The reversal of a number means the number formed by reversing the digits of the number, and for any number ending in 0 the reversal will not be defined or referred to in the puzzle. A number and its reversal are referred to as a reversal pair. A composite number is one which is not a prime.

Also: NDD refers to the number of digits in the (completed) diagram. NDD is NOT one of the 22 clue numbers.


(1) Cube root of the third largest number.
(2) Cube whose reversal is a prime.
(3) The only number having all square digits.
(4) Larger member of a reversal pair of triangles.
(5) Product of a palindrome and NDD.
(6) Number whose digits are all equal.
(7) Prime whose reversal is a cube.
(8) Product of a triangle and a palindrome.
(9) Product of four consecutive composite numbers.
(10) Composite reversal of a 4-digit prime.
(11) Smaller member of a reversal pair of triangles.
(12) Square of a palindrome.
(13) Ten less than a prime.
(14) Ten more than a prime.
(15) The only composite number whose digital sum is 22.
(16) The only triangle whose digital sum is 24.
(17) The smallest of four primes.
(18) The smallest of five triangles.
(19) The smallest of three palindromes.
(20) The sum of all of the digits in diagram columns 2,3,4,5 (left to right).
(21) Triangle which is also the sum of the squares of all of the digits in the diagram.
(22) Composite number which is two less than the largest palindrome.

Answers by PM please!

[dantuck_7]

ROUND 153

http://www.thesnookerforum.com/photo...ndex.php?n=275


Hello all, I'll close Round 153 - had a correct answer from snookersfun and a good guess from monique.

The image is a type of mathematical image known as Ulams prime spiral. If you start numbering from the centre and spiral outwards:

16 15 14 13
5 4 3 12
6 1 2 11
7 8 9 10

Then highlight any prime numbers you end up with the above pattern. It isn't as random as would be expected. Continue up until about 160,000 and you will get this image.

http://mathworld.wolfram.com/images/gifs/primesp.gif


The image that I posted on this thread starts numbering from the center at 41 and has primes all along the diagonal up until 1681.

Dan.

[davis_greatest]

Round 158 - Photo shoot

"While in Sheffield recently," I told my pet gorilla Gordon, "I noticed that the 32 finalists all differed slightly in height (some being more slight than others). I happened to take a photo of them all standing in line, and when editing it on my PC, I noticed that I could airbrush out 26 of them and leave the remaining six all in order of height. (The order could be either increasing or decreasing, and the six remaining were not necessarily standing next to each other.)"

"Fascinating," Gordon replied.

"What's more," I continued, "I had taken a number of shots, with the 32 players being in different orders each time, and soon realised that I could always remove 26 of them and leave six standing in height order - no matter what the original order of the players!"

"Well," said Oliver, my pet orang-utan, "Charlie is currently at the World Chimpanzee Snooker Chimpionships, and I bet that if you line up all the chimps there and take a photo, I'll be able to airbrush out some of them and leave 147 chimps all in order of height (which might be increasing or decreasing - again, the chimps might not necessarily be adjacent)."

"Bet you can't!" I said (rather stupidly, for Oliver's maths & logic skills always comfortably beat my own). "You don't even know yet in what order all the chimps will be standing for the photo!"

"I don't need to," said Oliver, "because I know how many chimps are in the World Chimpanzee Snooker Chimpionships!"

Well, sure enough, we took a photo of all those chimps standing in a long line, and Oliver studied it long and hard. But, after spending many hours on it, he finally admitted "I can't do it! I can't find 147 of them to leave in order of height! Someone must be missing!"

And then we looked, long and hard, at all those chimps, and finally saw why Oliver was having such problems! Charlie, my naughty pet chimpanzee, had been busy climbing my banana tree and was the only one of all the chimps who failed to get in the photo!

How many contestants in the current World Chimpanzee Snooker Chimpionships?

Congratulations to snookersfun and Monique for solving this! Both have now got there (At least, they got the answer, even though the proofs were somewhat dodgy .)

Answer and explanation are below in hidden text (so anyone who wants to keep trying can do so without seeing the answer) - select the text to read it.

(
There are 146² + 1 = 21317 chimps in the World Chimpanzee Snooker Chimpionships, but only 146² = 21316 made the picture, as naughty Charlie was missing!

We just need to show that
(a) if there are n² or fewer chimps, it is not always possible to select n+1 from them who are in order of size, and
(b) if there are more than n² chimps, it is always possible to select n+1 from them who are in order of size.

Then the answer follows when n=146.

Let's number the chimps in increasing order of size, from 1 upwards.

To show (a):
Imagine there are n² chimps lined up in groups of n, as follows:
n, n-1, n-2, n-3, ..., 1, 2n, 2n-1, 2n-2, ..., 2n-(n-1) , ..... , ...., n², n²-1, n²-2,..., n² - (n-1)

For example, if n were 4, we would have:

4,3,2,1, 8,7,6,5, 12,11,10,9, 16,15,14,13

We easily see that there can be no subgroup of n+1 increasing / decreasing chimps.


To show (b):
Imagine there are n²+1 chimps and suppose that:
(*) there is some arrangement of them for which it is not possible to select n+1 from them who are in order of size.

We will show that (*) leads to a contradiction.

For each chimp 1, 2, ... ,n²+1, assign him a pair of coordinates (x, y) such that, for that chimp:
x is the length of the longest sequence of increasing chimps, beginning with that chimp; and
y is the length of the longest sequence of decreasing chimps, beginning with that chimp.

For instance, if there are 5 chimps, in order 2, 5, 1, 3, 4
then for chimp 2,
x would be 3 (as the longest increasing sequence beginning with chimp 2 is {2,3,4} of length 3)
and y would be 2 (as the longest decreasing sequence beginning with chimp 2 is {2,1} of length 2)

For chimp 5, x would be 1 and y would be 2.

Now, if statement (*) were true, then 1<=x<=n and 1<=y<=n for every x and y.
Statement (*) therefore implies that there are only n² pairs of different coordinates (x,y).

Since there are n²+1 chimps, there must be two chimps i and j with the same coordinates, x_i = x_j and y_i = y_j.

Let's let chimp i be the one that comes before chimp j in the row (viewed from left to right). We know that either i
Now, if i x_j, a contradiction.

If i>j, then take chimp i and add the sequence of y_j decreasing chimps beginning with chimp j. This will give a decreasing sequence of chimps, beginning with chimp i, and length 1+y_j. So y_i > y_j, a contradiction.

Thus (*) leads to a contradiction and there must be a subgroup of n+1 chimps in order of size.

This completes the proof.

)