Puzzles with numbers and things

[davis_greatest]

Congratulations, snookersfun and robert602!

Oliver and Charlie are playing some more Chimpsnooker. They're back to a triangle with 20 rows of reds again (210 reds) and Gordon is refereeing and has set up the table. Once again, Gordon has set up the triangle upside down!

So Oliver and Charlie decide to remove some reds, to form a triangle the right way up. Once again, the triangle left can be any size (even as small as just one red!), and it doesn't matter whether it is shifted to the left or right, just as long as the apex red is towards the pink rather than towards the black!

So, a bit like round 97, how many possible triangles can be left, the right way up?

Answers by Private Message please

Congratulations, snookersfun and robert602, who both found that the answer to round 98 is 825. (robert602 was after the deadline but his answer is still being accepted)

This can be found by adding every second triangular number, up to the 20th.

The triangular numbers are:
1
1+2 = 3
1+2+3 = 6
1+2+3+4 = 10
1+2+3+4+5 =15
1+2+3+4+5+6 = 21
.....
1+2+3+...+19+20 = 210

If we add every second one we get 3 + 10 + 21 + ... + 210 = 825.

snookersfun will, I hope, paste a nice picture of triangles to explain why this is the solution. (abextra, where are you with your smiley triangle pictures for this round? )

For n rows, we can find a nice formula: for n even, the formula is n(n+2)(2n+5)/24, which, if we put n=20, gives the 825 above.

[davis_greatest]

He presses the buttons, then stops. And then takes a nap.

It's not really a puzzle. It's more a joke. So laugh. ;-)

Of course it's a real puzzle.... and a good one... so good that I'm thinking of awarding davis_greatest a centenary point.

[elvaago]

He presses the buttons, then stops. And then takes a nap.

It's not really a puzzle. It's more a joke. So laugh. ;-)

[davis_greatest]

...presses dozens and dozens and dozens of times. In fact, he presses the button seventeen thousand five hundred and thirty nine times!

That's a lot of dozens!

one in a hundred (assuming that he would have stopped for the nap after that time regardless of the colour of the light)

Edit: and assuming that by "done pressing the buttons" you mean done before the nap, and not on his return after the nap (otherwise the chance would be certainty)

[elvaago]

Round 100.

This round is worth no points!

I have a monkey. He's not very bright. In fact, all he can do is press buttons. and look at the wall. That's it. He's also colour blind. The only colour he can see is red. So he really likes red!

So I bought him a machine. The machine is basically a remote control for a beamer that's set in the ceiling. The beamer emits a big coloured circle on the wall. Every time he presses the button on the remote control, the colour of the beamer changes randomly. There is a one in a hundred chance that the colour turns to red. When it turns to red, my monkey will stop pressing the button.

Having said all that, the monkey starts pressing the button like a maniac. He presses dozens and dozens and dozens of times. In fact, he presses the button seventeen thousand five hundred and thirty nine times! Then he finally stops and decides to take a nap.

What are the chances of the light being red when my monkey is done pressing the buttons?

Answers in this thread.

[davis_greatest]

ROUND NINETY-NINE - Cubes of triangles!

You'll like this

Gordon has got lots of boxes, and Charlie has got loads of balls. Gordon likes cubes, and Charlie likes triangles.

Now, Gordon's boxes are all different sizes - in fact, he has one box of almost every size imaginable (but all perfect cubes). His smallest box is just large enough to hold one snooker ball. Every next box he has is one ball's width wider than the previous box. So his 2nd smallest box is two balls' wide (i.e. it can hold 2x2x2 = 8 snooker balls); his next box can hold 3x3x3 = 27 snooker balls etc.

Charlie lays his red balls out in a triangle and then Oliver, who has lots of golden balls, does an exchange. Oliver will give Charlie golden balls in exchange for each of Charlie's red balls - and the number of golden balls that Oliver will offer for each red ball is equal to the number of red balls in Charlie's triangle!

For example, if Charlie's triangle contains 15 red balls, then Oliver will offer Charlie 15 golden balls for each red ball, so Charlie would end up with 15 x 15 = 225 golden balls!

Well, they play this merry game, and then Charlie puts his newly-acquired golden balls into Gordon's boxes - starting by filling the smallest box, then the next smallest etc, until all the golden balls are in boxes. It turns out that all of Gordon's boxes are filled completely - there is just enough space for all the golden balls!

Now, I forgot to mention - Charlie did not start with a triangle of 5 rows (15 reds). In fact, his triangle had over a million rows! Hehe

If Charlie's initial triangle had had one row more than it did, and they had played this game, then how many extra boxes would Gordon have needed?

Answers initially by Private Message please

[davis_greatest]

Still only one correct answer to round 98. You can all have until 8pm GMT tonight, and then the round will be closed - so we can move on with new questions!

[davis_greatest]

Round 98 - Upside down Chimpsnooker

Oliver and Charlie are playing some more Chimpsnooker. They're back to a triangle with 20 rows of reds again (210 reds) and Gordon is refereeing and has set up the table. Once again, Gordon has set up the triangle upside down!

So Oliver and Charlie decide to remove some reds, to form a triangle the right way up. Once again, the triangle left can be any size (even as small as just one red!), and it doesn't matter whether it is shifted to the left or right, just as long as the apex red is towards the pink rather than towards the black!

So, a bit like round 97, how many possible triangles can be left, the right way up?

Answers by Private Message please

[davis_greatest]

Thanks to everyone for all the positive ratings! The rating still isn't back to where it was before (will need a lot of 5s to get there, but it's a start ).

Find a number, that by moving its last digit to the start of the number (e.g. 2349 becomes 9234) will result in an integral multiple (>1) of it.

i.e. abc...de*x=eabc...d

The smallest such number will earn a point. Answers, with explanations (not programs or excel) on the thread.

My answer to this was 102564, if I remember rightly. Take the 4 off the end and put it on the beginning, and you get 410256, which is an integer multiple (4x) of the original number. I'll allow snookersfun to give the explanation of how to get it!

Another possibility would have been the well-known 142857 (the recurring digits in the decimal expansion of 1/7 = 0.142857142857142857...) - move the 7 from the end to the start and you get 714285, which is 5 x the original number. (because 5/7 = 0.714285714285714285...)

But 102564 is smaller!




SO HERE IS THE SCOREBOARD AFTER ROUND 97

snookersfun.........................48
abextra...............................31
davis_greatest.....................24½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................11½
Sarmu..................................8
robert602.............................8
The Statman.........................5
austrian_girl and her dad.........3½
Semih_Sayginer.....................2½
Snooker Rocks! .....................2½
Ginger_Freak.........................2½
April Madness........................1

ROUND 98 ...

... later!

[davis_greatest]

1539 is the correct answer (Edit: to round 97, not to snookersfun's question!). You can solve this by counting the number of possible reds that could be at the top of each triangle, for each different sized triangle.

For a triangle of 20 rows (although this one is disallowed), there is just one red that can be at the top: the single red in the top row of the original triangle: 1

For a triangle of 19 rows, the red at the top can be any of the reds in the top 2 rows of the original triangle: 1+2 = 3

For a triangle of 18 rows, the red at the top can be any of the reds in the top 3 rows of the original triangle: 1+2+3 = 6

For a triangle of 17 rows, the red at the top can be any of the reds in the top 4 rows of the original triangle: 1+2+3+4 = 10

....

For a triangle of 1 row, the red at the top can be any of the reds in the top 20 rows of the original triangle: 1+2+3+4+5+...+20 = 210

If you add all these up, we are adding the first 20 triangular numbers. This is just like Snooker Rocks!'s question about the 12 Days of Christmas. If you put ever decreasing triangles on top of each other, you get a tetrahedron, and the nth tetrahedral number is n(n+1)(n+2)/6

Putting n=20, we get 20 x 21 x 22 / 6 = 1540. Or, you can add up all those triangular numbers by hand!

Now remember to subtract one because the original triangle of 20 rows is not allowed. So we get 1539.

Congratulations to snookersfun, robert602 and Ginger_Freak. Snooker Rocks! I think would have got there too with a little more time.

Scoreboard will follow later...

[snookersfun]

My deadline passed btw., so whenever the next points are doled out, d_g will get one point for my round and hopefully explain as well

[Ginger Freak]

Sorry - nursing a hangover from last night! Forgot you cant have the 1 triangle with all 210 balls!

1539?

[davis_greatest]

I see what you have done now, Ginger_Freak. You tripped at the last hurdle... and then turned back and tripped over it again. Just turn around again - you are nearly there!

I've had correct answers by Private Message from snookersfun and robert602...

[davis_greatest]

Oh yes - 1330!

Hehehe - you were indeed closer before!

[snookersfun]

Oh yes - 1330!

ouch..... and I thought you are there, lol