Puzzles with numbers and things

[elvaago]

There are N monkeys, so there are ((N-1)+(N-2)+..+(1)) number of matches. Let's call this number of matches M.
Each match gives out X bags of copper per ape and 1 bag of cold. So each match gives out 2X+1 bags. There are Y coins in each bag. Each match therefore gives out Y(2X+1) number of coins. There are M matches therefore they give out a total of MY(2X+1). This total equals 781153.

The equasion you need to solve is: 781153 = M*Y*(2X+1)

781153 is only dividable by 67, 89, 131, 5963, 8777 or 11659. M cannot be one of those. So M has to be 1. Therefore, there are only two apes. If there are 67 coins per bag, there are 5928 bags per ape per match. If there are 89 coins per bag, there are 4388 bags per ape per match. If there are 131 coins per bag there are 2981 bags per ape per match.

[snookersfun]

I think I followed it until this bit.... ? What are your h5, h6, l4?

we are moving ahead again, great

Concerning the above, I don't know, I just gave the light and heavy balls labels again (I thought it might make it clearer). Would have, if I had put the labels from the start, I guess (like put l1,l2,l3,h5,h6 vs 9,10,11,12,l4 ). Just assumed that the left side of the first weighing showed lighter balls (of course arguing would be similar for the heavy case).

Anyway, no other solutions, D_G has another point!

[davis_greatest]

There are N monkeys, so there are ((N-1)+(N-2)+..+(1)) number of matches. Let's call this number of matches M.
Each match gives out X bags of copper per ape and 1 bag of cold. So each match gives out 2X+1 bags.

I haven't read this all yet, but let my clarify if I was not clear:

1) The bags contain coins made of copper or of gold. Not cold. That would not be a very nice prize.

2) Each ape gets some bags of copper coins for competing in the competition - NOT for each match he plays.


And now I'll read the rest...

[davis_greatest]

Therefore, there are only two apes.

That would be the most boring league in the world. I can tell you that Gordon, Oliver and Charlie, at least, all play in the league.

[elvaago]

Darn. And I really thought I nailed it this time! Took me an hour too!

[davis_greatest]

Apart from that, elvaago, there's some good stuff in your answer. I'm tending towards giving you half a point.

[elvaago]

I misread the line about X bags of copper per ape, making it NX bags of copper and M bags of gold, where M is (n-1)+(n-2)+..+1.

So seeing as there are now NX bags of copper and M bags of cold... the equasion is MY+NXY=781153, where Y is still amount of coins per bag.

This makes my next answer: 67 apes, 56 bags of copper per ape.
N = # of apes = 67
M = # of matches = 2211
X = # of bags per ape = 56
C = # bags of copper = 3752
G = # bags of gold = 2211
T = # bags total = 5963
Y = # coins per bag = 131

[davis_greatest]

elvaago, so much work here, that I think I'll give you a full point for this.

But... it is not a full solution. It only shows that there COULD be 67 apes. Not that there must be. So, I will award:

one further full point to the first person who proves that there MUST be 67 apes in the league;

unless that first person is elvaago, in which case elvaago will get only an additional half-point.

[snookersfun]

for the proof hopefully:
the amount of coins can be also written as:
n(n-1)y/2 + nxy = 781153
(with n= number of players, y= number of coins per bag, x= number of bags with copper coins per player) thus the first part of the addition for number of gold coins won/game and second part for number of copper coins/player.

Therefore n and y need to be divisors of 781153, such that a number >1 remains. (possible 67,89,131).

Trying possible combinations, one has to remember, that the following have to be fullfilled:

(n-1)/2 +x = remaining factor (with n replaced by f1 or f2), e.g. with n=67:
x= 89-33=56, but n=131: x= 89-65=24
also n*x has to be divisible by 7, e.g. 56x67=3752=7*536, but 24*131=3144 doesn't divide by 7.

The equations solve only for n=67, y=131, yielding x=56 + all the other results, see elvaago (well done!)

[davis_greatest]

The equations solve only for n=67, y=131

Good try - again, most of the work done, but I don't think it's true that that is the only solution. I think I'll close the question there and post a new one soon. The catch was that there must be 67 apes, but we cannot work out conclusively how many coins are in each bag. Here is my answer.

Suppose there are:

k coins per bag
N apes
each ape gets j bags of copper coins for competing

There are (N-1) + (N-2) + (N-3) + … + 2 + 1 = N(N-1)/2 matches,
so N(N-1)/2 bags with gold coins, so k N (N-1)/2 gold coins in total

There are N j bags of copper coins, so k N j copper coins in total.

Total number of coins is:

k N (N-1)/2 + k N j = k N x [ (N-1)/2 + j ]
= 781153 (ESIIBLE upside down on a calculator display)

The prime factors of 781153 are 67, 89 and 131,
i.e. 781153 = 67 x 89 x 131.

Since k, N and [(N-1)/2 + j] are each greater than 1, each of them must take one of these 3 values.
(Alternatively, k, N/2 and N-1+2j could each have taken one of these 3 values, but that would not work since none of the 3 numbers 67, 89 and 131 is more than twice any other number and so it would give a negative j - therefore, we can discount that possibility.)

We know that the number of bags of copper coins, N j is divisible by 7 (since each day of the week appeared on each bag the same number of coins).
None of 67, 89 or 131 is divisible by 7, since those numbers are prime, so N is not divisible by 7.
Therefore j is divisible by 7.

If N is one of these three numbers (67, 89 or 131), say A; and (N-1)/2 + j is another number, say B, then

B = (N-1)/2 + j = (A-1)/2 + j

so

j = B-(A-1)/2 must be divisible by 7

Looking at the 6 possibilities, this is only the case if N = A = 67.

Therefore, there are 67 apes, and that is the answer to the question. We can stop here.



If you want to continue, you can observe that

(N-1)/2 + j = B can take either value 89 or 131.

So j can be either

89-(67-1)/2 = 56

or

131-(67-1)/2 = 98


There are 67x(67-1)/2 = 2211 bags of gold coins.

There is no way of knowing whether there are

67 x 56 = 3752 bags of copper coins, each containing 131 coins
or
67 x 98 = 6566 bags of copper coins, each containing 89 coins

Either way, there are 781153 coins (gold and copper) in total, and 67 apes playing.

[elvaago]

You're a strict teacher, davis_greatest. Here's an apple. ;-)

[davis_greatest]

Congratulations davis_greatest, Semih_Sayginer, elvaago and snookersfun

HERE IS THE SCOREBOARD AFTER ROUND 38, after awarding:

1 point to davis_greatest for round 37 (rambon's question about the 12 balls)
½ point to Semih_Sayginer for his treatise on battleships
1 point to elvaago for round 38
½ sympathy point to snookersfun for the semi-proof to round 38

We now have someone on the scoreboard for every day of the week.

snookersfun……………………….…..16½
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................2
Semih_Sayginer.....................½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[snookersfun]

Therefore, there are 67 apes, and that is the answer to the question. We can stop here.

I was too lazy to check all the options through, after the 67 was clear... should have, I guess

lol conc. Semih

[davis_greatest]

Round 39 - Ball tampering

Today is the first day of the last-16 matches of the Snooker Grand Prix on Astronomicus. Charlie, my pet chimpanzee, has travelled to Astronomicus, but has decided to cause a bit of mischief before the matches start, by mixing up the balls.

Snooker on the planet Astronomicus is rather similar to snooker on Earth, except it is played with more balls. The reds are formed in a triangle, like Earth snooker, but instead of there being 5 rows of reds (15 reds), the number of rows in the triangle is a 147-digit number. In fact, the number of rows is

14714714714714714714714714714714714714714714714714 71471471471471471471471471471471471471471471471471 47147147147147147147147147147147147147147147147
(yes, that is the digits 147 appearing 49 times).

It takes a lot of balls to play snooker on Astronomicus.

Before the eight last-16 matches start, Charlie takes the 8 triangles of reds from the 8 tables, and puts them in a very, very long line on Table Number 1 and places the cue ball at the end. So there are, say, N balls altogether in line (N-1 reds at the start, and then one white on the end).

On Table Number 2, Charlie places N-1 white balls and then one red ball at the end, ready for some mixing up fun. Then, he writes on bits of paper the numbers 2, 3, 4,… up to N, and puts them in a very, very big hat.

Gordon, my pet gorilla, who has travelled all that way with Charlie, is now invited to take a number out of the very, very big hat. Gordon draws the number 4, so Charlie switches every fourth ball on Table 1 with one in the equivalent position on Table 2. That means, counting from the start, he switches balls 4, 8, 12, 16, etc.

Then Gordon pulls out another number. It is number 75. So Charlie switches balls 75, 150, 225, … between the 2 tables.

They continue this game for a long time, until every single number has been pulled out the hat. Suddenly, it is time for the matches to start.

When Charlie and Gordon have finished, what is the colour of the last ball on Table 1 (where the cue ball was originally)? And how many red balls remain on Table 1?

[elvaago]

Excel blew up when I put in 14714714714714714714714714714714714714714714714714 71471471471471471471471471471471471471471471471471 47147147147147147147147147147147147147147147147. There's smoke coming out of my start menu!

I'll give this problem a lot of thought. It seems more philosophical than mathematical because the numbers are, pardon the pun, astronomically high!