Puzzles with numbers and things

[davis_greatest]

That is the correct answer! One very slightly different way to think of it would be to start by considering the numbers 0 to 999,999,999 (1,000,000,000 numbers).

If these all had 9 digits, there would be 9 x 1,000,000,000 digits = 9,000,000,000

But:

- the first 100,000,000 numbers don't have a digit in the 9th column from the right (the "hundred millions"), so subtract 100,000,000

- the first 10,000,000 numbers don't have a digit in the 8th column from the right (the "ten millions"), so subtract 10,000,000

- the first 1,000,000 numbers don't have a digit in the 7th column from the right, so subtract 1,000,000

... etc until we get to

- the first 10 numbers don't have a digit in the 2nd column from the right (the "tens"), so subtract 10

- the first 1 number (0) doesn't have a digit in the 1st column from the right (the "units") as it isn't on the list at all, so subtract 1.

We are left with

9,000,000,000 - 111,111,111 = 8,888,888,889.

Now add 10 digits for the number 1,000,000,000 and get 8,888,888,899


I need to decide whether still to award 1 point to anyone who submits the correct answer by PM by the Initial Deadline

[The Statman]

Very close - but not close enough for Gordon

1,000,000,000 digits in the units column,
same in the tens column, minus nine,
same in the 100s column, minus 99...

so it's the total of the following:

1,000,000,000
1,000,000,000-9
1,000,000,000-99
1,000,000,000-999
...
1,000,000,000-999,999,999

which is:

1000000000
999999991
999999901
999999001
999990001
999900001
999000001
990000001
900000001
1

which is 8,888,888,899.

??

[davis_greatest]

Well surely it's 9+(2×90)+(3×900)+(4×9000)+...+(9×900000000)+1 which is 8,888,888,890

Very close - but not close enough for Gordon

[The Statman]

But Gordon doesn't want me just to write the numbers from 1 to 15... he wants me to write the numbers 1 to 1,000,000,000

Well surely it's 9+(2×90)+(3×900)+(4×9000)+...+(9×900000000)+1 which is 8,888,888,890

[davis_greatest]

Latest scoreboard

I will shortly post snookersfun's round 66, slightly reworded!

elvaago and snookersfun, so far, have sent me correct answers to round 67 (although there is a neater method than that found by either of them!). Round 67 is still open for anyone else to have a go.

HERE IS THE SCOREBOARD AFTER ROUND 65 BUT BEFORE ROUND 66 AND WITH POINTS INCLUDED SO FAR FOR ROUND 67 FOR ELVAAGO AND SNOOKERSFUN

snookersfun……………………….…..31
abextra……………………………..…...16
Vidas……………………………………….12½
davis_greatest…………………..……12½
robert602…………………………………6
elvaago...............................6
The Statman……………………..……3
chasmmi ………………………………….3
Semih_Sayginer.....................2½

[davis_greatest]

Round 67 clarification

In response to a question I have received, I have asked Gordon to clarify. He wants to know the number of digits I would need to write - not the number of different digits.

For instance, if he wanted me to write the numbers from 1 to 15, I would need 21 digits:

1
2
3
4
5
6
7
8
9
1 0
1 1
1 2
1 3
1 4
1 5

Count them - there are 21 digits here. I had to use more than my fingers and toes to count that high.

But Gordon doesn't want me just to write the numbers from 1 to 15... he wants me to write the numbers 1 to 1000000000

[April madness]

I think you should tell Gordon to go to bed at this late night hour I'll give him a good night kiss if he wants to

[davis_greatest]

Round 67 - Not enough digits

Here is round 67, to run at the same time as snookersfun's round 66!

Gordon is teaching me some advanced maths. "Write out the numbers 1 to 10," he tells me. "How many digits do you need to write?"

I decide to count the digits out on my own digits - I start using my fingers, and find that I haven't got enough, so use one of my toes.

"11," I reply (realising that the 11 digits I need are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0).

"Well done, davis_greatest!" says Gordon.

"And how many digits do you need to write out the numbers 1 to 1,000,000,000?" he asks me.

I start counting out on my fingers, toes, and anything else I can find, but soon run out. I am stumped. What should I tell Gordon?


Answers by PM please, so Gordon doesn't know I have asked for help. Initial Deadline is 21:00 GMT on Wednesday 29 November.

[davis_greatest]

I have another try at rewording:
Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them, mark them and here is the important part, return them to the bag. At the best case scenario (choosing only unmarked balls) it is clear that one could mark all balls after 40 turns. But picking the balls randomly (or blind) how many balls would be probably marked after 40 turns and how many turns would one need to mark 90% of the balls.
Excel solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-12:00 pm (GMT)

This, of course, is round 66.

Um I'm still not sure what is being asked

1) What is meant by "probably marked after 40 turns"? Do you mean what is the average (i.e. "expected") number of balls that would be marked after 40 turns?

2) To mark 90% of the balls, you would need at least 90% x 40 turns = 36 turns! But I'm not going to let anyone get a point for that

[abextra]

Smaller Ape Break

red
black into black pocket
red
pink ....... green ........
red
blue ....... blue ..........
red
black ..... yellow ........
red
pink ....... pink ...........
red
blue ....... green .........
red
black ...... black .........
red
pink ........ green ........
red
blue ........ blue ..........
red
black ...... yellow ........
red
pink ........ pink ...........
red
blue ........ green ........
red
black ...... black .........
red
pink ........ green ........
red
blue ........ blue ..........
yellow ..... yellow .......
green ...... brown .......
brown ..... green ........
blue ........ blue ..........
pink ........ green ........
black ....... black ........

total break 154 points

[abextra]

Bigger Ape Break

red
pink into pink pocket
red
green ... green .......
red
black .... black .......
red
blue ...... yellow .....
red
pink ...... pink ........
red
green ... green .......
red
black .... black .......
red
blue ..... yellow ......
red
pink ..... pink .........
red
green ... green .......
red
black .... black .......
red
blue ..... yellow .......
red
pink ...... pink .........
red
green ....green ........
red
black .... black .......
yellow .. yellow ......
green ... brown .......
brown ... green .......
blue ...... blue ........
pink ...... green .......
black ..... black .......

total score 199 points

[snookersfun]

I have another try at rewording:
Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them, mark them and here is the important part, return them to the bag. At the best case scenario (choosing only unmarked balls) it is clear that one could mark all balls after 40 turns. But picking the balls randomly (or blind) how many balls would be probably marked after 40 turns and how many turns would one need to mark 90% of the balls.
Excel solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-12:00 pm (GMT)

This, of course, is round 66.

[snookersfun]

Here are the two max breaks that I found:

round 63) 15 reds-15 points

1)pink-pink-12
2)green-green-6
3)black-black-14
4)blue-yellow-5
5)pink-pink-12
6)green-green-6
7)black-black-14
8)blue-yellow-5
9)pink-pink-12
10)green-green-6
11)black-black-14
12)blue-yellow-5
13)pink-pink-12
14)green-green-6
15)black-black-14
=143 points

yellow-yellow-4
green-brown-3
brown-green-4
blue-blue-10
pink-green-6
black-black-14
=41 points

together: 199 points

round 64:
15 reds-15 points

1)black-black-9
2)blue-yellow-5
3)pink-pink-8
4)black-green-7
5)blue-blue-7
6)pink-green-6
7)black-black-9
8)blue-yellow-5
9)pink-pink-8
10)black-green-7
11)blue-blue-7
12)pink-green-6
13)black-black-9
14)blue-yellow-5
15)pink-pink-8
=106 points

yellow-yellow-4
green-brown-3
brown-green-4
blue-blue-7
pink-green-6
black-black-9
=33 points

together: 154 points!

[davis_greatest]

an easy one for in between...

Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them and mark them. At the best case scenario (choosing only unmarked balls) one could mark all balls after 40 turns, but how many balls would be actually marked after those 40 turns and how many turns would one need to mark 90% of the balls.
Excel solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-12:00 pm (GMT)

I'll PM you my questions on this, because I don't think I can have understood it correctly... if I have, it it is far too easy

[davis_greatest]

an easy one for in between...

Suppose you had a bag with 600 red snooker balls and would at each turn randomly take 15 of them and mark them. At the best case scenario (choosing only unmarked balls) one could mark all balls after 40 turns, but how many balls would be actually marked after those 40 turns and how many turns would one need to mark 90% of the balls.
Excel solutions worth 1/2 point, others 1 point. (by PM)
Deadline Tuesday-12:00 pm (GMT)

This, of course, is round 66.