Puzzles with numbers and things

[abextra]

update: I have two correct answers already by d_g and moglet. Well done!

...and Mon just sent a correct answer to R.380, despite high fever. Very well done and back into bed now!

Good job everybody, I'm really impressed! What was this 6-digit number?

Get well soon, Monique!

[davis_greatest]

Good job everybody, I'm really impressed! What was this 6-digit number?

Get well soon, Monique!

I don't remember! I never remember or keep the answers! (Well, is probably in my sent PMs.) I'm sure moglet will have it though.

Get well soon, Monique.

snookersfun, moglet, Monique and abextra all solved round 381 Snap happy. Would someone please put up the answer, with a mini-explanation.

[moglet]

R380

Good job everybody, I'm really impressed! What was this 6-digit number?

Get well soon, Monique!

Speedy recovery Monique.

Abextra, I think R380 is still open.

[snookersfun]

Good job everybody, I'm really impressed! What was this 6-digit number?

Abextra, I think R380 is still open.

Should I close this round then?
In that case the number and explanations can go up on the thread now

[abextra]

Should I close this round then?
In that case the number and explanations can go up on the thread now

Good, I would like to see the explanation...

[davis_greatest]

I just found another good one (should probably credit some javaguru for those):

R.380 more primes in 6-digit numbers

This time Gordon tells Charlie he's thinking of a 6-digit number. All its digits are different. The digital sum of the number is equal the number formed by the last 2 digits in the number. Moreover the sum of the first 2 digits is the same as the sum of the last 2 digits.
If one takes the sum of the number, the number formed by rotating its digits one to the left, the number formed by rotating digits one to the right, the number formed by rotating digits pair-wise to the right, the number ‘pair-wise rotated’ to the left, and the number with the first three and last three digits swapped, one can observe that the first and last digits of this sum match the last two digits of the number, in some order.
Gordon concludes: ‘If each of the three numbers formed by the digit pairs in the number is prime, then what is my number?’
Charlie has a little ‘think’ and comes up with which number?

Answers by PM please

Yeah, I would like to see the explanation... couldn't understand the question again, sorry.

The digital sum means the sum of the digits - e.g. digital sum of 147 is 1+4+7 = 12.

Unfortunately I don't remember the answer and wiped my PMs, but hopefully we can do it again. So...

The 6 digits of the mystery number are all different, so they must sum to at least 0+1+2+3+4+5=15 but no more than 4+5+6+7+8+9=39.

This sum, being the digital sum, is equal to the number formed by the last two digits, which is prime - so the only possibilities for these last two digits in the range 15 to 39 are: 17, 19, 23, 29, 31 and 37.

We can immediately discount four of these, by remembering that the sum of the first 2 digits equals the sum of the last two:

- try 17 - sum of these digits is 8, so sum of first 2 digits is also 8, leaving sum of middle 2 digits to be 17-8-8 = 1 --- impossible (since the middle 2 digits would need to be 01 and we have used 1 already)

- try 19 - sum of these digits is 10, so sum of first 2 digits is also 10, and these already sum to 20 which is > 19 --- impossible

- try 31 - sum of these digits is 4, so sum of first 2 digits is also 4, leaving sum of middle 2 digits to be 31-4-4 = 23 --- impossible

- try 37 - sum of these digits is 10, so sum of first 2 digits is also 10, leaving sum of middle 2 digits to be 37-10-10 = 17 --- meaning middle 2 digits must be 89 (since they must form a prime number)... but then only possibilities for first 2 digits are 19 or 91 (no - used 9 already and 91 isn't prime), 28 or 82 (no -not prime), 37 or 73 (no - used 3 and 7 already), 46 or 64 (no - not prime), 55 (no - uses same digit twice) .... impossible

This leaves possibilities for last 2 digits to be 23 or 29.

Note that adding all the rotations of the 6-digit number gives 111111 x sum of its digits, i.e. adding
ABCDEF +
BCDEFA +
CDEFAB +
DEFABC +
EFABCD +
FABCDE

gives A+B+C+D+E+F in each column

Therefore, we can readily rule out 29 for the last 2 digits, since 111111 x 29 clearly starts with 3 and ends in 9, so does not use the digits 2 and 9.

This leaves the last 2 digits being 23. These sum to 5. So the first 2 digits must also sum to 5, and the only prime that works would be 41 (05 not allowed since the number can't start with 0).

Thus the middle 2 digits must sum to 23 - 5 - 5 = 13, and 67 is the only 2-digit prime whose digits sum to 13.

So the number - hopefully - was 416723.


Now someone can put up the answer to round 381?

[davis_greatest]

Round 382 - Blazing Mad

The Gorilla Team is up against the Orang-utan Team (each Team having the same number of apes) in the Great Ape Snooker-Wrestle-Off. The match will consist of numerous rounds. Each round consists of a Subteam of gorillas against a Subteam of orang-utans.

A Subteam can be any size from one ape up to the whole Team.
For example, if the Gorilla Team happened to consist of 3 gorillas (A,B & C) - say - then the possible Gorilla Subteams would be A; B; C; AB; AC; BC; and ABC.

Every possible Gorilla Subteam plays exactly one round against each possible Orang-utan Subteam! There doesn't need to be an equal number of apes in each Subteam. (So, in the example above, there would be 7x7= 49 rounds.)

They keep score in a rather unusual, some might say stark crazy, way. The score is kept by the winning Subteam of each round taking pieces of baize. At the start, a large rectangular piece of baize had been stripped from a snooker table and laid out. Then, each gorilla had folded it once in half from left to right, and each orang-utan had folded it in half from top to bottom – alternating between gorilla and orang-utan until every single ape had made one fold.

Then the still-folded cloth was cut all the way along with two perpendicular cuts – one vertically from top-middle to bottom-middle; and one horizontally from left-centre to right-centre. This left many, many pieces of baize.

Now, every time the gorillas win a round, they take one piece of baize; and every time the orang-utans win, they take two pieces! It just happens that the orang-utans win 128 rounds in total. It also happens that, by the end of the match, every single piece of baize is taken, with there being exactly the right number needed to keep score!


So, how many apes were there in each Team?

Answers by Private Message please.
Bit of a confusing one, this. So, any questions, requests for clarification or requests for hints, please ask here.

[snookersfun]

Round 382 - Blazing Mad

[COLOR="DarkGreen"]They keep score in a rather unusual, some might say stark crazy, way. The score is kept by the winning Subteam of each round taking pieces of baize. At the start, a large rectangular piece of baize had been stripped from a snooker table and laid out. Then, each gorilla had folded it once in half from left to right, and each orang-utan had folded it in half from top to bottom – alternating between gorilla and orang-utan until every single ape had made one fold.

Answers by Private Message please.
Bit of a confusing one, this. So, any questions, requests for clarification or requests for hints, please ask here.

yes, a bit Are we to ignore that one can really only fold a limited number of times?
goes to cut paper now

and :
thank you for putting up the solution for R.380, very well done and congratulations again to d_g, moglet and Mon

[davis_greatest]

yes, a bit Are we to ignore that one can really only fold a limited number of times?

Yes, the apes are very strong and they can fold the baize as many times as they like. Also, this baize is rather old and has worn thin, so that helps!

[moglet]

R381

I hope this will show the routine, A,B,C,...P represent the original queue order, the downward arrows show that this player went to the table and when, the other arrows show that this player was sent to the rear of the queue. Player A went first and is ranked 1st, player B went 7th and is ranked 7 and so on. The eeny meeny miny moe shuffling after the eighth player is difficult to show but players E and L at the end had to swap places 14 times. Very easy to make mistakes and get confused

[davis_greatest]

I hope this will show the routine, A,B,C,...P represent the original queue order, the downward arrows show that this player went to the table and when, the other arrows show that this player was sent to the rear of the queue. Player A went first and is ranked 1st, player B went 7th and is ranked 7 and so on. The eeny meeny miny moe shuffling after the eighth player is difficult to show but players E and L at the end had to swap places 14 times. Very easy to make mistakes and get confused
[ATTACH]2008[/ATTACH]

Thank you moglet, and congratulations again to you, snookersfun, Monique and abextra on solving round 381. One easy way to do it, not too dissimilar from the method you have shown, is to leave 16 rows for the numbers and write 1 in the first row, then move down 2 blank places and write 2, move down 3 blank places and write 3; etc. Also:

a) whenever you need to go down below the bottom row, start again at the top row (or you could write the numbers in a circle if you prefer); and

b) whenever you find there is already a number there in a row as you count down, you skip over it to the next blank space.

This you continue until you have written all the numbers from 1 to 16.

--------------

R382 update - congratulations to snookersfun, who is the first to send a correct answer. I am happy to give any hints to anyone who may want any. That is particularly important on this question, as I regret that I can accept no liability to anyone who strips and cuts up any baize from any snooker table while working on this round.

[davis_greatest]

R382 update - congratulations to Monique, abextra and moglet who have joined snookersfun in solving this. Would someone please put up a solution...

[Monique]

R382 solution

lets suppose we have a gorillas and b orans:
- therefore we have (2^a -1) gorilla teams and (2^b -1), hence (2^a -1)*(2^b -1) matches played.
- a gorillas produce 2^a -1 folds marks in the baize, dividing it in 2^a "zones" (in one direction, say in the lenght) that when "cut" would produce 2^a +1 "pieces"; similarly b orans would produce 2^b +1 "pieces" ; as they fold in perpendicular directions the combination will produce (2^a +1)*(2^b +1) pieces.

So we have (2^a -1)*(2^b -1) matches played and (2^a +1)*(2^b +1) pieces taken, meaning that the difference between these two numbers is 2^(a+1) + 2^(b+1) and is also equal to the number of rounds won by the orans because they take two pieces for each victory, so 128 = 2^(a+1) + 2^(b+1)

Now because we have a=b (as I had overlooked was stated in the "story", but still proved to be the only possibility!) 2^(a+2) = 128.

Hence a=5=b, 5 gorillas and 5 orans.


PS an elagant way to compute the number of possible teams if starting with n apes is to assign a "position" number to each ape from 1 to n. Then for each "team" an ape gets a "1" token if part of it and a "0" token if not. So each team is associated with exactly one number represented in binary format and all values are met from 1 to 2^n -1 ...

[davis_greatest]

Round 383 - Ninety nine Valentine

Oliver is on his way out tonight and intends to give a rose to each of his orang-utan lady friends. Each rose costs ninety pence, and he intends to pay using exactly 9 coins for each rose. He also intends to use a different combination of coins for each purchase.

Up to how many lady orang-utans could be receiving a nice surprise from Oliver tonight?

Bids can go here on the thread!

British coins in common circulation - and the only ones to which Oliver has access - are: 1p, 2p, 5p, 10p, 20p, 50p, £1, £2

[abextra]

Round 383 - Ninety nine Valentine

Up to how many lady orang-utans could be receiving a nice surprise from Oliver tonight?

Bids can go here on the thread!

At least 13.