Puzzles with numbers and things

[davis_greatest]

Question 40 - penny squares

thanks, good puzzle, but.... what some people do in bed....

Indeed! Well, this next one wasn't conceived in bed - but at my desk at work just now So, here goes.

Charlie now takes the last row of reds from a triangle of reds from a game of snooker on Astronomicus (as described in the previous question) - i.e. 14714714714714714714714714714714714714714714714714 71471471471471471471471471471471471471471471471471 47147147147147147147147147147147147147147147147 reds. (That should hopefully be a 147-digit number.)

Each red ball costs 37 Astronomical pence.

He places these reds on a big Astronomicus chessboard, the same number of reds in each square. Chess on Astronomicus, like on Earth, is played on a square board, but does not necessarily have 64 squares. All I know is that it does not have more than 100 squares.

Charlie then works out, in Astronomical pence, the total cost of reds in each square on the chessboard (i.e. the cost in one square). He then adds up all the digits of this number.

What is his answer for the sum of the digits?

(Note - Charlie did not use a calculator or computer for this, so you are not allowed to either. )

[Semih_Sayginer]

i dont understand! could you repeat the question please?

(wouldnt mind a job in your work too. seems easy)

[davis_greatest]

i dont understand! could you repeat the question please?

(wouldnt mind a job in your work too. seems easy)

As an example, if there were 50 reds (instead of that 147-digit number), then Charlie might put them on a 5x5 chessboard, 2 reds in each of the 25 squares.

Each red costs 37 Astronomical pence, so the 2 reds in each square would come to a total of 2 x 37 Astronomical pence = 74 Astronomical pence.

Charlie would then add up the digits, 7+4 and get 11.

But, there are not 50 reds...

[snookersfun]

any chance for 520?

[davis_greatest]

any chance for 520?

Not what I made it when I used my fingers... but you can set out your reasoning and see if you can persuade me?

[Semih_Sayginer]

d_g i think ive got it now. you want an answer?

btw, i noticed im on the points list. 1/2 a point for half a brain eh?

keep up the good work. i like this thread

[snookersfun]

I take that back then, 147, better?

[davis_greatest]

Semih, let's hear your answer then! Hold fire for a moment, snookersfun - let's give him a chance to build on that half-point!

[Semih_Sayginer]

588

hey i know its wrong, but at least i smiled

[April madness]

btw, i noticed im on the points list. 1/2 a point for half a brain eh?

I think you might be right here. At least my amount of brain perfectly corresponds to my point count on the scoreboard

[davis_greatest]

588

hey i know its wrong, but at least i smiled

Afraid not, Semih...

I was thinking about this while going home last night and remembered that I had not stated, as I had intended, that the chessboard must be bigger than 9 squares. Oops. As I forgot to say this, this means that there are two possible solutions - a chessboard of 9 squares (3x3) or a chessboard of 49 squares (7x7).

So, snookersfun's two answers of 520 and 147 were, indeed, both possible! The answer I had been looking for (assuming the chessboard must be bigger than 9 squares) was a 7x7 chessboard with a sum of digits of 147.


With a 7x7 chessboard, the reds on each square cost a total of 11111.....111 (147 "1"s) Astrononomical pence. So, adding the 147 "1"s gives a total of 147.


One point to snookersfun, and also one point to abextra who paged me the answer of 147.


Here is snookersfun's answer, sent to me by Private Message:

147 seems to be the easy answer for a chess-board with 49 squares. each
147 unit of the big number of course divides nicely by 49 to yield 3, which multiplied by 37 gives 111, thus the final result will be a number 111...111 again 147 digits, therefore sum of 147.

But, I think my first answer of 520 is OK as well, although it will need a bit more paper (as you don't want a calculator). This time it is a rather small chessboard as well (9 squares). Each 147147147 unit now divides by 9 and multiplied by 37 gives a sum of digits of 40. As there are 13 such units, one should get a sum of 13*40=520 for this case. I only calculated one unit yesterday by hand, but checked for two units today with the big number calculator, and it seems to hold.

Possible chess boards, of course were 4,9,25,36,49,64,81,and 100. None of the even ones can be divisors, as your number is not even. Further 25 not possible, as number doesn't end in 5 or 0. 81 wouldn't work, as an even number of 147147147 fragments would be needed. Leaving 49 and 9.

Scoreboard to follow...

[davis_greatest]

HERE IS THE SCOREBOARD AFTER ROUND 40:

snookersfun……………………….…..18½
Vidas……………………………………….8½
abextra……………………………..…...6½
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)

[davis_greatest]

Round 41 - Metal medals and kiss kiss

Let's see how challenging you find this one! It took me long enough to come up with it (although most of my deliberations were in trying to make up a decent story ) ....



On Astronomicus, as you know, snooker is very popular. In fact, there are very, very many players, most of whom like to play for the medals that they can win. All medals are the same size, but come in different metals.

The format is a knock-out - all players enter in the first round.

At the start of the Astronomicus Snooker Grand Prix last week, the organising committee (my pet orang utan Oliver and my pet gorilla Gordon) worked out how to reward the players.

They decided that everyone who entered the competition would be given one bronze medal for taking part.

Everyone who reaches the halfway stage will then be given a silver medal as well (e.g., if you need to win 10 matches to win the tournament, then when you have won 5 you would have reached the halfway stage; but as you will soon see, a player needs to win a lot more than 10 matches!).

After being awarded a silver medal, if a player wins a further 147147 matches, he is also given a gold medal to go with the bronze and silver he has already! That's a lot of rounds to get through!

And after being given a gold medal, a player would have to win another 147147 matches just to reach the final! Yes, as you can see, there are many, many competitors, and so many, many rounds.

Winning the Astronomicus Grand Prix is therefore quite an achievement. In fact, the winner will be given a platinum medal.

Before the tournament starts, Oliver and Gordon decide it will make good publicity to have photos of the medals taken by the media. So they start arranging all the medals in a big rectangle.

In the top row, in a long line, they place all the silver and gold medals and the platinum medal.

Underneath, they start placing the bronze medals in rows (each row the same length of the top row, otherwise it wouldn't be a very good rectangle). However, when they get to the last row, they find that they don't have enough bronze medals to complete the row!

"That won't do," cries Gordon. "It won't look good for the photo if the medals don't fill the last row. What can we do?"

"I know," says Oliver. "Let's make a medal out of iridium to fill each remaining space in the last row. Then we can give an iridium medal to everyone who wins a certain minimum number of matches."

"How many matches would someone need?" asks Gordon.

"Work it out!" says Oliver. "It's easy!"

"Ah yes, of course it is." says Gordon. "Silly me!"

"OK," says Oliver. "And every time someone who has won an iridium medal loses a match, you have to give him or her a kiss!"


And so, your question: how many kisses does Gordon have to give?

[snookersfun]

I won't even have time to read that now, as this is my last check before getting ready for my plane.
All of you have some happy puzzeling next week.

[abextra]

I'm afraid there will be no kisses from Gordon, at least not in this tournament.