Puzzles with numbers and things

[davis_greatest]

Round 136 - Kiss me!

Did anybody solve the "question"?

What question? And why in quotes? Was there something wrong with it?

I indeed did receive correct answers to round 135 from snookersfun, rambon and abextra. They are all already in the Hall of Frame but have cemented their positions even more firmly - well done! If anyone else (or one of these three) wants to post an answer on this thread, please feel free.

Meanwhile, moving on - here is a true (almost) story…

Round 136 - Kiss me!

Last night, Gordon and Oliver played some snooker. Gordon was so good that Oliver was left to sit and watch for most of the evening, doing a brilliant impression of John Parrott during the 1989 World Final, and in fact he amused himself by playing with his recent Valentine's Day presents (a whole load of daffodils from his new girlfriend Olivia, a romantic present of a cannon from Charlie, and a whole load of equally romantic cacti from me).

Actually, they only got as far as playing one frame, with Gordon knocking in a cool 147 maximum directly from Oliver's otherwise excellent break-off. Every red that Gordon potted, he played as a plant (as long as there was more than one red on the table). On the first plant that he played, and every 2nd one thereafter (so on the 1st, 3rd, 5th…), he also played a cannon into others reds or colours. However, every time that he played a cannon after having potted a black on the previous shot, he got a double kiss. This didn't bother him though and he remained in perfect position, aided by his impressively long arms and brown fur.

The main reason that they only managed one frame during the evening was due to Oliver's antics, him playing the fool somewhat. Oliver handed Gordon a congratulatory daffodil every time that Gordon played a plant, except when Gordon also played a cannon on the same shot, in which case Oliver instead put a plant (either a daffodil or cactus) into his cannon and fired it at Gordon's face. During the frame, Gordon was hit in the face by more cacti than by daffodils (only by one more though, fortunately), but continued his break unperturbed.

Every time that Gordon was handed a daffodil, he gave Oliver a kiss. Whenever Gordon played a double kiss, Oliver gave him a double kiss. Whenever Gordon was hit in the face by a cactus, he gave Oliver a treble slap. Whenever Gordon was hit in the face by a daffodil, he gave Oliver a hug. Each time that Oliver was slapped, he cried and Gordon had to give him a quadruple kiss to cheer him up. Whenever Gordon gave Oliver a hug, Oliver handed Gordon a daffodil.

During the evening, there were no plants, cannons, cacti, daffodils, roses, geraniums, hugs, slaps, kisses, double kisses, squeezes, quintuple massés, sextuple girlfriends, fouls, fowls, foals, or fools, Chinese snookers or 18-3 thrashings of poor John Parrott other than those already described above.

How many kisses, hugs and slaps were there, and how many points did Oliver score?

[rambon]

OK, I'll have a go at asking a question...

At the recent Chimpanzees World Snooker Chimpionship, 256 chimps competed in the televised knock-out stages at The Chimpible.

1) What was the average number of matches that each chimp won?
A) none; B) between 0 and 1/2; C) 1/2; D) between half and 1; E) 1; F) more than 1; G) can't say

255 games required, so the average wins/chimp is 255/256, so D

2) What proportion of chimps won a match (i.e. at least one match)?
A) none; B) between 0 and 1/2; C) 1/2; D) between half and 1; E) 1; F) can't say

Half lose in the first round so don't win any games. Therefore C

3) What proportion of chimps lost a match?
A) none; B) between 0 and 1/2; C) 1/2; D) between half and 1; E) 1; F) can't say

Only 1 chimp can win the tournament, and he will be the only one not losing so it's 255/256 again, or D

4) What proportion of chimps lost more matches than they won?
A) none; B) between 0 and 1/2; C) 1/2; D) between half and 1; E) 1; F) can't say

Only those that lose in the first round, therefore 128 out of 256 or half, C

5) If you write out the answers to questions 1-4 as exact numbers (rather than letters), how many different answers would you have?
A) 0; B) 1; C) 2; D) 3; E) 4; F) can't say

2: 255/256 twice and 1/2 twice

6) If you write out the answers to these 6 questions as exact numbers (rather than letters), how many different answers would you have?
A) 0; B) 1; C) 2; D) 3; E) 4; F) 5; G) 6; H) can't say

We have 2, 255/256 and 1/2. We so far have 3, but if we answer 3, that means we have 4 different answers, so it means this part would be wrong. The answer therefore is 4, as that makes us have 4 different answers....

[davis_greatest]

Correct, correct, correct, correct, correct and correct!

[davis_greatest]

3 correct answers to round 136... from snookersfun, April Madness and abextra... oh dear, have I missed someone, or was that it? Anyway, someone please post the answer on this thread (one of these three or anyone else who wants to) - I've had enough of this question... perhaps we can move on?

What question? And why in quotes? Was there something wrong with it?

I indeed did receive correct answers to round 135 from snookersfun, rambon and abextra. They are all already in the Hall of Frame but have cemented their positions even more firmly - well done! If anyone else (or one of these three) wants to post an answer on this thread, please feel free.

Meanwhile, moving on - here is a true (almost) story…

Round 136 - Kiss me!

Last night, Gordon and Oliver played some snooker. Gordon was so good that Oliver was left to sit and watch for most of the evening, doing a brilliant impression of John Parrott during the 1989 World Final, and in fact he amused himself by playing with his recent Valentine's Day presents (a whole load of daffodils from his new girlfriend Olivia, a romantic present of a cannon from Charlie, and a whole load of equally romantic cacti from me).

Actually, they only got as far as playing one frame, with Gordon knocking in a cool 147 maximum directly from Oliver's otherwise excellent break-off. Every red that Gordon potted, he played as a plant (as long as there was more than one red on the table). On the first plant that he played, and every 2nd one thereafter (so on the 1st, 3rd, 5th…), he also played a cannon into others reds or colours. However, every time that he played a cannon after having potted a black on the previous shot, he got a double kiss. This didn't bother him though and he remained in perfect position, aided by his impressively long arms and brown fur.

The main reason that they only managed one frame during the evening was due to Oliver's antics, him playing the fool somewhat. Oliver handed Gordon a congratulatory daffodil every time that Gordon played a plant, except when Gordon also played a cannon on the same shot, in which case Oliver instead put a plant (either a daffodil or cactus) into his cannon and fired it at Gordon's face. During the frame, Gordon was hit in the face by more cacti than by daffodils (only by one more though, fortunately), but continued his break unperturbed.

Every time that Gordon was handed a daffodil, he gave Oliver a kiss. Whenever Gordon played a double kiss, Oliver gave him a double kiss. Whenever Gordon was hit in the face by a cactus, he gave Oliver a treble slap. Whenever Gordon was hit in the face by a daffodil, he gave Oliver a hug. Each time that Oliver was slapped, he cried and Gordon had to give him a quadruple kiss to cheer him up. Whenever Gordon gave Oliver a hug, Oliver handed Gordon a daffodil.

During the evening, there were no plants, cannons, cacti, daffodils, roses, geraniums, hugs, slaps, kisses, double kisses, squeezes, quintuple massés, sextuple girlfriends, fouls, fowls, foals, or fools, Chinese snookers or 18-3 thrashings of poor John Parrott other than those already described above.

How many kisses, hugs and slaps were there, and how many points did Oliver score?

[snookersfun]

round 137

Round 137

I found a nice puzzle
Here it goes (I trust you all not to google for the solution):

The Island of Apes and Monkeys
On a remote island all of the natives belong to one of two tribes: the Apes, who are so brilliant at numerical calculations that they always get the correct answer, and the Monkeys, who bravely rush in to do calculations beyond their ability and never get the right answer. (The Monkeys are not entirely stupid: they can do simple counting and comparing of numbers, but they always get arithmetic calculations wrong.) Both Apes and Monkeys pride themselves on their complete honesty. They always tell the truth, or (in the case of Monkeys) at least what they believe to be the truth; they never purposely tell a lie (unlike the folks on some of those other islands).
One day a group of natives was playing a game of Numberskulls. There were 5 players and a moderator. The moderator, who was an Ape, painted a 3-digit number on each of the players' foreheads, so that each could see all numbers but their own. All 5 of the numbers were different. The moderator would ask them questions in turn about the numbers they could see, and from the answers they would try to deduce what number was on their own forehead. The first to do so was the winner. What follows is a record of the game, with questions omitted and players designated by letters.

(1) A: I see exactly 1 prime number.
(2) B: I see exactly 2 prime numbers.
(3) C: I see exactly 3 perfect squares.
(4) D: I see exactly 3 triangular numbers.
(5) E: I see exactly 3 perfect squares.

(6) A: I see exactly 3 numbers with a digital sum of 10.
(7) B: I see exactly 3 numbers whose square root is more than 25.
(8) C: I see exactly 0 numbers with a digital sum of 10.
(9) D: I see exactly 3 perfect cubes.
(10) E: I see exactly 0 numbers with a digital product of 18.

(small tip: if a Monkey gives an answer 'I see x properties' they rather see (4-x) properties)

At this point one of the players announced his number and won. (Of course it was an Ape; for some reason Monkeys never win these games, a point of much amusement to the Apes!)


What number was on each player's forehead?

Answers by PM please!

[abextra]

Round 136 - Kiss me!

Last night, Gordon and Oliver played some snooker. Gordon was so good that Oliver was left to sit and watch for most of the evening, doing a brilliant impression of John Parrott during the 1989 World Final, and in fact he amused himself by playing with his recent Valentine's Day presents (a whole load of daffodils from his new girlfriend Olivia, a romantic present of a cannon from Charlie, and a whole load of equally romantic cacti from me).

Actually, they only got as far as playing one frame, with Gordon knocking in a cool 147 maximum directly from Oliver's otherwise excellent break-off.

It makes 0 points to Oliver.

Every red that Gordon potted, he played as a plant (as long as there was more than one red on the table)On the first plant that he played, and every 2nd one thereafter (so on the 1st, 3rd, 5th…), he also played a cannon into others reds or colours. However, every time that he played a cannon after having potted a black on the previous shot, he got a double kiss. This didn't bother him though and he remained in perfect position, aided by his impressively long arms and brown fur.

So he played 14 plants, 7 cannons (on shots 1, 3, 5, 7, 9, 11, 13) and got 6 double kisses (on shots 3, 5, 7, 9, 11, 13).

The main reason that they only managed one frame during the evening was due to Oliver's antics, him playing the fool somewhat. Oliver handed Gordon a congratulatory daffodil every time that Gordon played a plant, except when Gordon also played a cannon on the same shot, in which case Oliver instead put a plant (either a daffodil or cactus) into his cannon and fired it at Gordon's face. During the frame, Gordon was hit in the face by more cacti than by daffodils (only by one more though, fortunately), but continued his break unperturbed.

It makes 7 daffodils handed, 4 cacti fired and 3 daffodils fired.

Every time that Gordon was handed a daffodil, he gave Oliver a kiss.

7 kisses

Whenever Gordon played a double kiss, Oliver gave him a double kiss.

6 * 2 = 12 kisses

Whenever Gordon was hit in the face by a cactus, he gave Oliver a treble slap.

4 * 3 = 12 slaps

Whenever Gordon was hit in the face by a daffodil, he gave Oliver a hug.

3 hugs

Each time that Oliver was slapped, he cried and Gordon had to give him a quadruple kiss to cheer him up.

12 * 4 = 48 kisses

Whenever Gordon gave Oliver a hug, Oliver handed Gordon a daffodil.

3 kisses

During the evening, there were no plants, cannons, cacti, daffodils, roses, geraniums, hugs, slaps, kisses, double kisses, squeezes, quintuple massés, sextuple girlfriends, fouls, fowls, foals, or fools, Chinese snookers or 18-3 thrashings of poor John Parrott other than those already described above.

How many kisses, hugs and slaps were there, and how many points did Oliver score?

So in total there were 7 + 12 + 48 + 3 = 70 kisses, 3 hugs, 12 slaps and 0 points to little Oliver.

[snookersfun]

Round 137

The Island of Apes and Monkeys
Answers by PM please!

A quick update on this. I just had a perfect solution PMed by Berolina. Well done!
I'll leave it open for a bit longer, so come on all the rest!

[rambon]

No time to do it at the moment... hears a similar one before though

[snookersfun]

A quick update on this. I just had a perfect solution PMed by Berolina. Well done!
I'll leave it open for a bit longer, so come on all the rest!

It seems to be high time to close this round. Abextra solved it perfectly as well! So, either Berolina or Abextra please put up your solution. If anybody else wants a go at it, he/she can put up the solution as well of course.

[abextra]

The Island of Apes and Monkeys.

(1) A: I see exactly 1 prime number.
(2) B: I see exactly 2 prime numbers.
(3) C: I see exactly 3 perfect squares.
(4) D: I see exactly 3 triangular numbers.
(5) E: I see exactly 3 perfect squares.

(6) A: I see exactly 3 numbers with a digital sum of 10.
(7) B: I see exactly 3 numbers whose square root is more than 25.
(8) C: I see exactly 0 numbers with a digital sum of 10.
(9) D: I see exactly 3 perfect cubes.
(10) E: I see exactly 0 numbers with a digital product of 18.

(small tip: if a Monkey gives an answer 'I see x properties' they rather see (4-x) properties)

Answer: A - 631
. . . . . . .B - 136
. . . . . . .C - 233
. . . . . . .D - 361
. . . . . . .E - 343

A was the only Ape and he won the game, of course.

Attempt to explain:

At first I checked the lists of numbers:

3-digit perfect squares - 100, 121, 144,169,196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961

3-digit perfect cubes - 125, 216, 343, 512, 729

3-digit triangular numbers - 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990

3-digit prime numbers - http://primes.utm.edu/lists/small/10000.txt

From (2) and the tip

a) B saw exactly 2 prime numbers

There are no similar numbers in the lists of triangular numbers and perfect cubes, no one could see 3 triangular numbers and 3 perfect cubes at the same time, so from (4), (9) and a tip

b) D was a monkey, he saw exactly 1 triangular number and 1 perfect cube.

From (6) and (8) it's obvious that A and C can't be the same native. If C was an ape, then from (3) he had to see 3 perfect squares, but from (1) and the tip A had to see 3 prime numbers at the same time. It's impossible, so

c) A was an ape, he saw 1 prime number and 3 numbers with a digital sum of 10.

d) C was a monkey, he saw 1 perfect square and 4 numbers with a digital sum of 10.

From a) there were 2 prime numbers, from b) there was 1 triangular number. There are no similar numbers in the lists of triangular numbers and perfect squares, so E couldn't see 3 perfect squares, as stated in (5). It means

e) E was a monkey, he saw exactly 1 perfect square and 4 numbers with a digital product of 18.

Now we know, that

there were 2 prime numbers, 1 perfect square, 1 perfect cube and 1 triangular number,

A, B, D and E had numbers with a digital sum of 10 on their foreheads and

A, B, C and D had numbers with a digital product of 18 on their foreheads.

From a) and c) A must have had a prime number on his forehead. There are 4 prime numbers with a digital product of 18 -163, 233, 613 and 631. 3 of them have a digital sum of 10, so

A's number was 163, 613 or 631.

There is only 1 perfect cube with a digital sum of 10 and none with a digital product of 18, so the perfect cube must have been on E's forehead:

E's number was 343.

B, C and D must all have had numbers with a digital product of 18. There is only 1 triangular number with a digital product of 18 - 136. It also has a digital sum of 10, it means it couldn't be on C's forehead and as D could see this number, it has to be on B's forehead:

B's number was 136.

C saw perfect square, so it must have been on D's forehead.There is only 1 perfect square with a digital product of 18 and a digital sum of 10:

D's number was 361.

C must have had a prime number with a digital product of 18 and it's digital sum wasn't 10. The only possibility is:

C's number was 233.

So far all numbers have been smaller than 625, so their square roots are smaller than 25. From (7) and a tip

B was a monkey and he saw exactly 1 number whose square root is more than 25.

A's number was 631.

Sorry for my English.

[snookersfun]

Abextra, well done, great explanation as well No worry about the English!

[snookersfun]

Round 138

moving on to the next round:

Round 138:

Archeologists recently made an amazing find in a new dig site not far from Rome. It was an approximately two thousand year old stone tablet on which was engraved what could only be described as a primitive cross-number diagram. It had only four numbers, two horizontal (reading left to right) and two vertical (reading top to bottom), as shown here in the blank diagram with the numbers removed. All four of the numbers were different and all were less than four thousand. The bottom horizontal one was the smallest. After consulting with mathematicians they discovered that all four of the numbers were squares, but only the smallest number was a fourth power. Also the sum of the four numbers was a prime. Using only the information given here can you find the unique original diagram?


Answers initially by PM please

[snookersfun]

Round 139

... and I am through another one to pass on before Curtis will cut us off.
This one is a bit easier:

Round 139:
Find a positive integer (base 10) which has the following four properties:

(1) Each of the 10 digits (0 to 9) appears exactly once in the number.
(2) For each pair of digits whose sum is 9, the number of other digits positioned strictly between the pair is equal to the smaller digit of the pair.
(3) The sum of each pair of digits positioned at the same distance from opposite ends of the number is a prime number.
(4) The difference between any 2 adjacent digits is greater than 1.

The solution is unique.

[snookersfun]

moving on to the next round:

Round 138:

Archeologists recently made an amazing find in a new dig site not far from Rome. It was an approximately two thousand year old stone tablet on which was engraved what could only be described as a primitive cross-number diagram. It had only four numbers, two horizontal (reading left to right) and two vertical (reading top to bottom), as shown here in the blank diagram with the numbers removed. All four of the numbers were different and all were less than four thousand. The bottom horizontal one was the smallest. After consulting with mathematicians they discovered that all four of the numbers were squares, but only the smallest number was a fourth power. Also the sum of the four numbers was a prime. Using only the information given here can you find the unique original diagram?


Answers initially by PM please

A tip concerning this one, I have just discovered that excel does have a ROMAN function, which should make live a lot easier!

[berolina]

Sorry! No time this week...been busy learning french, 6 hrs per day, still got no clue, bl**** grammar.