Puzzles with numbers and things

[snookersfun]

@Dan, yeah, that helped a whole lot! Round-robin indeed

regarding my rounds, I won't close them yet, let's give others a chance at cracking them as well for a little bit more...

and meanwhile, in case anybody is bored during sessions, I'll put up another crossword (don't let yourselves be put off by the larger numbers, they are quite easy to figure out once you'll take a closer look). So:

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!

[davis_greatest]

Congratulations to both snookersfun and Monique for solving round 150, which is now closed. Well done (and apologies for my delay in confirming it here)! Here is Monique's picture below (snookersfun submitted the same answer).

I shall also invite those hosting all questions but the last open round to close their rounds, so that we avoid having too many open rounds (and so I can resume posting questions )...

[snookersfun]

closing rounds

with the promise of more puzzles looming on us, I shall speedily close some rounds:
Round 151 and Round 152:
solved by d_g and Monique. Congrats and well done, especially on the tricky round 151

I invite those two to post their answer to puzzle 151!

solution to round 152 up here, it was an easy one (all other possibilties would have a 0 starting a number and therefore be invalid, as all numbers 1-9 appear as starting numbers in the 'crossnumber')

[davis_greatest]

Round 155 - Out of my league

I play snooker in a league, and during the season each person plays every other person once. Always the same number of frames are played each match (with "dead frames" being played out), so that a player might win, lose or draw. In fact, we have arranged it so that the number of frames in a match is the same as the number of players in our league!

At the end of the season, while looking down the results, I was amazed to find that every possible match score appeared the same number of times. (A score of x-y is deemed here to be the same as a score of y-x.)

Just like the number of this round - being the largest possible break in snooker - the number of players in my league was the greatest possible size for the above to occur.

How many play in my league?

[davis_greatest]

Congratulations to Monique for correctly identifying that there can only be 4 players in my league in round 155. The scores 4-0, 3-1 and 2-2 each appeared twice, during the 6 matches.

Round 156 - Time for more Grottslaps

Raymond Grott has just finished his first season playing in the Exciting Snooker League. In the Exciting Snooker League, each player plays every other one once. Unfortunately, Grott loses every one of his matches, and no one ever turns up to watch him play. Each time that Grott loses, he whines and accuses his opponent of slow play or cheating. Each time that Grott whines, his opponent gives him a slap. Each time that an opponent slaps Grott, all the other players in the league buy tickets to come and watch. In fact, each time, the slapper enjoys it so much that he pays for a ticket himself.

At the end of the season, Grott's matches / slapping sessions have netted £1471.47 in ticket sales.

How many players in Grott's league, and how much is each ticket?

[davis_greatest]

Correct answers to round 156 received from snookersfun and Monique - well done! I'll leave the round open a little while longer, for anyone else who wants to have a go.

But as the World Championships are on, things will move quickly - so anyone else trying it, get a move on, as here's the next question...

Round 157 - Pool party

Last night, my pet chimpanzee Charlie took me to a pool party. There were 20 other chimpanzees there, each of whom had also brought a human friend.

At the party, some of us (chimpanzees and humans) had met others there before. (Of course, each chimpanzee knew the human he had brought.) Each time someone met someone that he/she had never met before, they played a frame of pool. The number of chimpanzees I played was two more than the number of humans I played (I beat all the humans I played but always lost to the chimpanzees, the cheeky apes).

At the end of the party, Charlie asked everyone at the party (including me) how many frames of pool he / she had played. Astonishingly, everyone gave a different answer.

How many frames did I win and how many did I lose?

[davis_greatest]

... and to prove it can be done, snookersfun has answered round 157 already! It's still open...

[davis_greatest]

... and to prove it can be done, snookersfun has answered round 157 already! It's still open...

...and Monique has too!

[davis_greatest]

OK, we'll soon close all previous rounds. Last chance if anyone else wants to try them!

In the meantime...

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?

[davis_greatest]

Round 156 - Time for more Grottslaps

Raymond Grott has just finished his first season playing in the Exciting Snooker League. In the Exciting Snooker League, each player plays every other one once. Unfortunately, Grott loses every one of his matches, and no one ever turns up to watch him play. Each time that Grott loses, he whines and accuses his opponent of slow play or cheating. Each time that Grott whines, his opponent gives him a slap. Each time that an opponent slaps Grott, all the other players in the league buy tickets to come and watch. In fact, each time, the slapper enjoys it so much that he pays for a ticket himself.

At the end of the season, Grott's matches / slapping sessions have netted £1471.47 in ticket sales.

How many players in Grott's league, and how much is each ticket?

Congratulations to snookersfun and Monique, for correctly identifying that there are 8 players in the league, each paying £30.03 for a ticket.

It can be solved as follows. Assume there are n players in the league, so n-1 play Grott, and after each match these n-1 players buy tickets, which we assume cost X pence each.

Then (n-1)(n-1)X = 147147

So we need to find a square factor of 147147

Now 147147 = 3 x 7 x 7 x 7 x 11 x 13

So (n-1) = 7
and
X = 3 x 7 x 11 x 13 = 3003

I.e. n=8

[davis_greatest]

Round 157 - Pool party

Last night, my pet chimpanzee Charlie took me to a pool party. There were 20 other chimpanzees there, each of whom had also brought a human friend.

At the party, some of us (chimpanzees and humans) had met others there before. (Of course, each chimpanzee knew the human he had brought.) Each time someone met someone that he/she had never met before, they played a frame of pool. The number of chimpanzees I played was two more than the number of humans I played (I beat all the humans I played but always lost to the chimpanzees, the cheeky apes).

At the end of the party, Charlie asked everyone at the party (including me) how many frames of pool he / she had played. Astonishingly, everyone gave a different answer.

How many frames did I win and how many did I lose?

Congratulations to snookersfun and Monique, for correctly identifying that I won 9 frames and lost 11, by realising that I must have played 20 frames.

They (or anyone else) are invited to post their explanations here.

[davis_greatest]

And on round 158 (3 posts up), congratulations to snookersfun, who has got there with the answer (and is closing in on a successful explanation / proof). I'll leave it open a little while longer for anyone else who wants a try - it really isn't that hard!

[Monique]

Solution to R157

D_G lost to 11 apes ... and won 9 humans

Here is the reasoning ...
First if A met B before, B obviously also met A before
Then Charlie who asked the question is the only one who didn't answer it ...

Charlie got 41 different answers, including D_G's, going from 0 to 40. Lets name In the individual, human or champ, who answered "n"

I40 didn't met anybody before except his "friend"
I0 met everyone, including I40 and therefore must be I40's friend

I39 met only one individual other than his friend before; this individual therefore must be I0
I1 knew everyone except one individual, I40 and hence must be I39 friend

You can continue along this recurrent line of reasoning, and conclude that friends come in pairs {Ik,I40-k} for k going from 0 to 20 ...

Now we have an interesting pair ... {I20,I20}; as there is only one I20 the other member of this pair can only be Charlie ... and D_G is I20!

[davis_greatest]

Thanks Monique - nice answer - so, any luck at round 158?

[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.

)