[Mentalist]

rofl
5016's a huge number to start with

surely there must be a smaller number?

[eekwibble]

It is not possible for them to wake up in the morning to the '4 piles' that have been left which will divide equally into 5 with 1 remainder.

It would require a number ending in a 6 since it must be divisible by 4 and be divisible by 5 with a 1 remainder (just expanding on your reasoning there Ment).

The only integers ending with a '6' that divide by 4 into integers (up to 100) are; 16, 36, 56, 76 and 96.

If the last guy left 16 he must've had 21 when he woke up. 21 doesn't divide into 4 so it can't be that.

If he left 36 he must've had 46 when he awoke and that doesn't divide into 4 either.

If it was 56 left he would've woken up to 71 (you can see that this doesn't equate).

If it was 76 it was 95...

...96 means 121... blah blah blah...

Riddlefail!!!

[Giles]

15621 works we think its trying to find a smaller number

[Sir Thursday]

The way I did it was to work backwards from the end. When they all wake up in the morning, they divide up 5n+1 coconuts.

Ergo (5n+1/4) must be an integer. In modular arithmetic, we can say that 5n=-1 (mod 4), which is equivalent to n= -1 (mod 4).

We also need (5*(5n+1/4)+1)/4 to be an integer. Expanding out, you get (25n+9)/16, so 25n=-9 (mod 16), which is equivalent to 9n = -9 (mod 16). So n= -1 (mod 16).

You will note a pattern developing here. Basically, you work your way up and find that n=-1 (mod 1024). The smallest positive integer with that property is 1023, and if you work through the problem you get 15621 coconuts as your final solution.


Sir Thursday

[Giles]

Yeah thats basically what i did to, I got the wrong answer yesterday as my formula didnt take into account needing integer values at each stage i think, i was just looking for integer values at the end.

[Mentalist]

eekwibble, on Feb 25 2009, 07:28 AM, said:

It is not possible for them to wake up in the morning to the '4 piles' that have been left which will divide equally into 5 with 1 remainder.

It would require a number ending in a 6 since it must be divisible by 4 and be divisible by 5 with a 1 remainder (just expanding on your reasoning there Ment).

The only integers ending with a '6' that divide by 4 into integers (up to 100) are; 16, 36, 56, 76 and 96.

If the last guy left 16 he must've had 21 when he woke up. 21 doesn't divide into 4 so it can't be that.

If he left 36 he must've had 46 when he awoke and that doesn't divide into 4 either.

If it was 56 left he would've woken up to 71 (you can see that this doesn't equate).

If it was 76 it was 95...

...96 means 121... blah blah blah...

Riddlefail!!!

actually, 56 would've meant 72, since you're forgetting to add the extra 1, and that's divisible by 4.
but it gets screwed up later up in the chain.
the quarter of the number has to end in a 9 for it to work, so it should always be a multiple of 36.
but yeah, I think we can agree that ST got it, and it's now up to him top post a new riddle.

[Sir Thursday]

Hmm, I think I'll go with a slightly easier, less logical one this time:

What cheese is made backwards?


Sir Thursday

[eekwibble]

Edam. I likes the easy ones.

The last one broke my brain.

[frookenhauer]

dammit I missed the easy one...riddle us quick eek

[eekwibble]

What grows down while it grows up?

@Ment - (56 x 1.25) + 1 = 71



Actually I've just looked at STD's answer to the previous riddle and I don't get why you guys think that it's possible for them all to wake up in the morning to any number that doesn't end in a '6'. It's fairly basic maths. The 5th guy left 4 stacks (an even number) that divide into 5 with a 1 remainder. It MUST end in a '6'!

This post has been edited by eekwibble: 26 February 2009 - 08:56 AM

[HoosierDaddy]

A root system?

[Cause]

A duck!

[eekwibble]

A duck or goose. Correct. Well done!

[frookenhauer]

riddle us quickick!

[Cause]

Sorry cant today. Frook cover me quick!

[frookenhauer]

Hot damn...

A man is buying a gold ring set with stones for his wife on her birthday. A ring set with 4 amethysts and 1 diamond comes to US$2,000. One with 3 emeralds, 1 amethyst and 1 diamond would be US$1,400. And one set with 2 rubies and 1 diamond would cost US$3,000. Being a thoughtful husband, he choses a ring with 4 stones, each representing one of their 4 children.

As their children are named Andy, David, Ellen, and Richard, how much in US$ will the ring containing 1 amethyst, 1 diamond, 1 emerald, and 1 ruby cost him?


Edit:

For those that do not like numbers here's a quickie....

You're sitting in a car that's not moving with a helium-filled balloon, which is resting up against the car's ceiling somewhere near its middle. The driver hits the gas and the car accelerates forward, throwing you back into your seat. What happens to the balloon (spoilerise your answers)

This post has been edited by Frookenhauer : 26 February 2009 - 07:23 PM

[Sir Thursday]

Well I reckon that the ring that the man wants would cost $1900.

From the second equation, we know that 1a + 1e = 1400 - 2e - 1d, and from the third we know that 1r + 1d = 3000 - 1r, so 1a + 1e + 1r + 1d = 4400 - (2e + 1r + 1d).

By comparing the first and third equations, we see that 1r = 2a + 500, and by comparing the first and second we see that 1e = 1a - 200. So we can rewrite the equation for the ring he wants as: 1a + 1e + 1r + 1d = 4000 - (2a + 500 + 2a - 200 + 1d)

Which becomes 1a + 1e + 1r + 1d = 4000 - (4a + 1d + 100)
And then: 1a + 1e + 1r + 1d = 4000 - (2100)

To give 1a + 1e + 1r + 1d = 1900.


Still thinking about the balloon.


Sir Thursday

[frookenhauer]

Sometimes I wish I was still doing maths...most days not. I'm sad about the most days part.

[Impirion]

STD was right with respect to my riddle, sorry for breaking other people's minds, I was just trying to break his

Is STD right wrt your puzzle Frookie? Because if he is, then being the lazy bugger I am, I will just wait for his riddle.

[frookenhauer]

Yup he was right with the numbers 1...forgot about his savant like abilities with number stuff...The balloon one is up for grabs...