Puzzles with numbers and things

[snookersfun]

counting 0 s though (I'm still a bit condused by the question), that should rather be: 17400

so zeros at the end, that are 'only' 1320

[snookersfun]

DG seems to be gone. And I hopefully arrived at a final answer for question 24:

There should be 1115 final 0s

Proof: 147 games with 147 maximums (36 balls each) were played, therefore 5292 balls were played
Calculations were done 147*1*2*3...*5292 = 5292!*147
Now each multiplication by a number that is a multiple of 5 (0 or 5 last digit) can yield a zero at the end of the result. Moreover whole hundreds yield 2 zeros and whole thousands 3 zeros, therfore
1058+52+5= 1115 zeros at the end of the crazy big result, (which I tried to calculate above and gave the powers of 10)

[snookersfun]

a) 48 golden balls
b) 30 bags (but here my math might forsake me again, prove will follow later as usual)
c) 1 ball into each back and remaining 18 balls into bag #15. Now, if (and I wonder why) the girls are 16,17,18 and 19 years old only the last one could pick up her one bag, while the others would not find a combination.

Although question 22 is still open, I'll set question 23 now before I go away for a few days. This one is intended to take a little longer to solve - but it's always hard to tell!

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Oliver, whom you will know very well by now as my pet orang-utan, goes on holiday to Krypton to watch a snooker tournament, which never stops. They play snooker there every day, just like snooker on Earth, except there is no "free ball" rule (so the maximum break is 147). Oliver arrives to see the first day's play, and stays there each day after that.

Every day, they flash up on a big screen the top century break that was made that day (if one was). Every time that the top century break is different from all the top century breaks that have come before, everyone in the audience is given a golden ball to keep as a souvenir.

Eventually, realising he will get no more golden balls, Oliver comes home.

Part a (easy) - how many golden balls does Oliver bring home?

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Then Gordon, whom you will also know very well by now as my pet gorilla, goes on holiday to Moominon to watch a snooker tournament, which never stops. He arrives at the start of the month in time for the start of the tournament.

On Moominon, they play snooker just like on Krypton, with no free ball, except that on Moominon the players are very good and every time that a player wins a frame, he always makes a break at least as high as the highest break made in the Final by the winner of the 2006 888.com World Snooker Championship (which I will be nice and tell you was 68).

Every day, the officials write up on a blackboard, in descending order, the top 5 breaks made that day. (More than 5 frames are played a day, so they can always find a top 5, each one at least 68.) For instance, they might write "134, 126, 103, 103, 76". (If there are two or more "5th-equal" breaks - so if there were more than one 76 break in this example - they would only write one of them, so the list would still appear as shown above.)

At the end of each month (which happens to consist of 1 million very quick days on Moominon), everyone in the audience is given a transparent goody bag. Oliver eats the goodies but keeps the bags as souvenirs.

Gordon likes to keep track of these daily lists of the top 5 breaks. It's incredible, but every day Gordon notices that the list is different from every previous day - i.e. never the same list twice! This goes on, until one day Gordon realises that if he stays another day, a list is bound to come up that will be the same as one of the ones he has seen before. So he comes home with his empty goody bags.

Part b (harder) How many empty goody bags does Gordon bring home?

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When they get home, Charlie, whom you will also know very well by now as my pet chimpanzee, lays out all of Gordon's empty goody bags in a long straight line on the floor. Into each, he puts at least one golden ball, until all of Oliver's golden balls are in bags.

(If you have more bags than balls, you need to go back and check your parts a and b.)

Then, Florence, Elizabeth, Sylvia and Talia (all of whom are aged younger than the number of goody bags) come to visit. Charlie tells them that if any girl can find a set of one or more adjacent bags, containing a total number of balls equal to her age, then she can keep those bags and the balls they contain. For example, if a girl is 10, and if there are 7 balls in bag 4, 1 ball in bag 5 and 2 balls in bag 6, she could take those 3 bags as the number of balls adds up to her age (10) and the bags are adjacent. In this example, if she were aged 7, she could simply take bag 4.

Any balls not taken will be kept by clever Charlie. He doesn't know the girls' ages but wants to keep as many balls for himself as he can.

Florence, the youngest, steps up, but sadly cannot find any bags that work, and goes home empty-handed. Then up steps Elizabeth, the next youngest. She leaves and goes home, just as disappointed. Then up comes Sylvia, older still, and she has no more success, so also goes home.

Finally, Talia, a year older than Sylvia, picks a bag and goes home with her balls.


Part (c) (not sure how hard you will find this):
How old is each girl, and how many golden balls does Charlie keep?


* "older" means at least one year older

[snookersfun]

a) 48 golden balls
b) 30 bags (but here my math might forsake me again, prove will follow later as usual)
c) 1 ball into each back and remaining 18 balls into bag #15. Now, if (and I wonder why) the girls are 16,17,18 and 19 years old only the last one could pick up her one bag, while the others would not find a combination.

no editing, so I will use my 400th post to correct slightly.
a) of course it is proof
b) you also asked, how many balls he keeps, so 48-19=29

[Vidas]

DG seems to be gone. And I hopefully arrived at a final answer for question 24:

There should be 1115 final 0s
Proof: 147 games with 147 maximums (36 balls each) were played, therefore 5292 balls were played
Calculations were done 147*1*2*3...*5292 = 5292!*147
Now each multiplication by a number that is a multiple of 5 (0 or 5 last digit) can yield a zero at the end of the result. Moreover whole hundreds yield 2 zeros and whole thousands 3 zeros, therfore
1058+52+5= 1115 zeros at the end of the crazy big result, (which I tried to calculate above and gave the powers of 10)

multiplies of 25, 125, 625 will yield more than one zero as well(2,3,4 resp.), because there are enough multiples of 2 for them. I would stick to 1320...

[snookersfun]

multiplies of 25, 125, 625 will yield more than one zero as well(2,3,4 resp.), because there are enough multiples of 2 for them. I would stick to 1320...

... and here are the missing zeros... you should get points for that!

[Vidas]

... and here are the missing zeros... you should get points for that!

i didn't find any solution - just picked the right answer from various
posts above , so it doesn't count.

Question 22

n1,n2,n3,n4,n5,n6,n7,n8,n9,n10 (10 numbers)
all sums of 2 numbers:
n1+n2, n1+n3,... n9+n10. (10!/2!8! = 45 numbers.)
all triples
n1+n2+n3, (120 numbers)
sums of 4 different numbers
n1+n2+n3+n4, etc. - 210
sums of 5 different numbers
n1+n2+n3+n4+n5, etc. - 252.
Total 10+45+120+210+252 =637 numbers,
all sums don't exceed 99+98+97+96+95=485.
so we can always find 2 equal sums there - and select 2 sets.

[davis_greatest]

I am back now! I see there have been some answers! I just need to read back and remind myself what questions I asked - and then see what points should be awarded...

[davis_greatest]

i didn't find any solution - just picked the right answer from various
posts above , so it doesn't count.

Question 22

n1,n2,n3,n4,n5,n6,n7,n8,n9,n10 (10 numbers)
all sums of 2 numbers:
n1+n2, n1+n3,... n9+n10. (10!/2!8! = 45 numbers.)
all triples
n1+n2+n3, (120 numbers)
sums of 4 different numbers
n1+n2+n3+n4, etc. - 210
sums of 5 different numbers
n1+n2+n3+n4+n5, etc. - 252.
Total 10+45+120+210+252 =637 numbers,
all sums don't exceed 99+98+97+96+95=485.
so we can always find 2 equal sums there - and select 2 sets.

Good answer, Vidas, but you forgot the most important bit - to answer the question and say who wins the game! The answer is that Oliver can always choose two sets of numbers that have the same sum, so he will always win. I'll give you the point though!

The easier way I was thinking that you would solve this - although on similar lines to the above - would be to say that there are 2^10 - 1 = 1024 - 1 = 1023 ways of choosing a set of numbers from the 10 numbers.

This is because for each number, there are two choices - include it or exclude it. So this gives 2^10 = 1024 choices. Then deduct the one way where no numbers at all are chosen, to arrive at 1023 choices.

The maximum sum is 99+98+97+96+95+94+93+92+91+90 = 945.

So, when choosing two of 1023 possible sets, each of which can add up to one of 1, 2, 3, 4, ..., 945, at least two of them must have the same sum. So Oliver can always win.

[snookersfun]

I am back now! I see there have been some answers! I just need to read back and remind myself what questions I asked - and then see what points should be awarded...

Welcome back!!!

[davis_greatest]

Question 24 - I'll be kind

DG seems to be gone. And I hopefully arrived at a final answer for question 24:

There should be 1115 final 0s

Proof: 147 games with 147 maximums (36 balls each) were played, therefore 5292 balls were played
Calculations were done 147*1*2*3...*5292 = 5292!*147
Now each multiplication by a number that is a multiple of 5 (0 or 5 last digit) can yield a zero at the end of the result. Moreover whole hundreds yield 2 zeros and whole thousands 3 zeros, therfore
1058+52+5= 1115 zeros at the end of the crazy big result, (which I tried to calculate above and gave the powers of 10)

This, snookersfun, as I think you probably know by now from your and Vidas's subsequent posts, isn't really quite right, but as you gave the correct answer of 1320 in the previous post (albeit without explanation) and as I'm just back from holiday, I'll give you the point! Well done. 1,320 is the correct answer.

There are 147 total clearances, each of 36 pots, so 147 x 36 = 5,292 pots.

WT starts with the number 147, and muiltiplies it by 1, 2, 3, 4, ..., 5292. So we need to find the number of zeroes at the end of the number 147 x 5292! (i.e. 147 x 1 x 2 x 3 x ... x 5292).

This means we need to find how many factors of 10 this number has, which is equivalent to counting the factors of 5 in 5292! (since factors of 2 appear more often than the factors of 5, and the initial 147 has no factors of 5).

Every 5th number, a factor of 5 appears - so 5292 / 5 = 1058 times (only count whole numbers)

Every 25th number (5^2) another factor of 5 appears - so 5292 / 25 = 211 times

Every 125th number (5^3) another factor of 5 appears - so 5292 / 125 = 42 times

Every 625th number (5^4) another factor of 5 appears - so 5292 / 625 = 8 times

Every 3125th number (5^5) another factor of 5 appears - so 5292 / 3125 = 1 time

The 5^6th number never appears.

So the number ends in 1058 + 211 + 42 + 8 + 1 = 1320 zeroes.

[davis_greatest]

Question 23 - one point already, snookersfun!

a) 48 golden balls
b) 30 bags (but here my math might forsake me again, prove will follow later as usual)
c) 1 ball into each back and remaining 18 balls into bag #15. Now, if (and I wonder why) the girls are 16,17,18 and 19 years old only the last one could pick up her one bag, while the others would not find a combination.

Well done, snookersfun. I need to decide how many points that is worth. Can you at least prove your 30 bags for part b?

(Part a is trivial - there are 48 possible breaks 100, 101, ...., 147)

[snookersfun]

It's worth lots of points. Let me find my scribbles.
Did you see the welcome-back post above. You seem to be deeply absorbed in the math!

OK, found something. It's messy, trying to extract:
I added up all the possible combinations:
1) w.o. repeated numbers (there are 80 possible different high-breaks possible): 80!/5!(80-5)! = 24.040.016
2) all #s the same = 80
3) 4 of the same #s = 80x79 = 6320
4) 3 of the same #s + 2 of same #s = 80x79 = 6320
5) 3 of same + 2 different #s = 80x79x78/2 = 246.480
6) 2 of same + 3 singles = 80x79!/3!(79-3)! = 6.326.320
7) 2 pairs + single like 5

sum was less than 31.000.000, therefore 30 bags

[davis_greatest]

Welcome back!!!

Thank you snookersfun!

[davis_greatest]

It's worth lots of points. Let me find my scribbles.
Did you see the welcome-back post above. You seem to be deeply absorbed in the math!

OK, found something. It's messy, trying to extract:
I added up all the possible combinations:
1) w.o. repeated numbers (there are 80 possible different high-breaks possible): 80!/5!(80-5)! = 24.040.016
2) all #s the same = 80
3) 4 of the same #s = 80x79 = 6320
4) 3 of the same #s + 2 of same #s = 80x79 = 6320
5) 3 of same + 2 different #s = 80x79x78/2 = 246.480
6) 2 of same + 3 singles = 80x79!/3!(79-3)! = 6.326.320
7) 2 pairs + single like 5

sum was less than 31.000.000, therefore 30 bags

Good one! You could have got there a bit faster by noting that the number of different lists (in descending order) of 5 breaks, each of which can take 80 different values, is the same as the number of ways of choosing 5 objects from 80+5-1=84 objects.

I.e. it is 84! / 5!(84-5)! = 30,872,016. So, with one bag being given every million days, there are 30 bags.

Definitely 2 points for this!