ofc, i was thinking to myself, it would be impossible with a maximum of 1999... so let me have another crack at it

Edit: my current answer is, 3,281 Watermelons in the crate

This is wrong.

click here  :kappa:

click here  :kappa:

the cow never dies. :doge:

Well since nobody posted the correct answer yet I'll give you guys a tip.

It's easier when you reverse things.

The best I can come up with is 3121. If that isn't it I surrender :v

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

You said there were 2 crates on the ship but only one crate was used in these series of events.

From there I can only speculate that the crates were identical and contained half half of the total 4000 watermelons. So 2000.

2000 is divisible by 5, so there wouldn't be a leftover melon for the monkey.

Anyway,

Quote from: Chazz on May 11, 2015, 07:08:17 pm

You said there were 2 crates on the ship but only one crate was used in these series of events.

From there I can only speculate that the crates were identical and contained half half of the total 4000 watermelons. So 2000.

2000 is divisible by 5, so there wouldn't be a leftover melon for the monkey.

This riddle is easy. the monkey used to much oil :kappa:

6 pls

Quote from: Ttk on May 23, 2015, 11:21:04 am

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

5^5 is 3125 and if each person has one left over then the number has to end in a 1 or 6
I tried it with 3126 but it didnt work so I tried it with 3121 and it did
3121 -1 /5 gave 624. if the first sailor hid it then there would only be 624*4 watermelons left.
repeated this till I got 255 on the fifth time(?). a lot of pain was involved c:

Quote from: Lars on May 23, 2015, 11:31:11 am

Quote from: Ttk on May 23, 2015, 11:21:04 am

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

5^5 is 3125 and if each person has one left over then the number has to end in a 1 or 6
I tried it with 3126 but it didnt work so I tried it with 3121 and it did
3121 -1 /5 gave 624. if the first sailor hid it then there would only be 624*4 watermelons left.
repeated this till I got 255 on the fifth time(?). a lot of pain was involved c:

Ah I haven't seen anyone use this approach, very creative.

I personally used a different approach when I first came across this problem. I started from the bottom and figured out that the final amount of coconuts has to be divisible by 5 and 4, in other words a multiple of 20. I also figured out, just like you that every amount of coconuts in the crate has a 1 or a 6 as it's last digit. However I also knew no amount could end with a 1 (except for the starting amount). Thus every upstream amount ends with 6, which means it cannon be a multiple of 40. This is because 40 * 1.25 + 1=51. Which means the final amount of coconuts is a multiple of 20, but not a multiple of 40. I found these limitations sufficient to start using trial and error.

Quote from: Ttk on May 23, 2015, 09:57:40 pm

Quote from: Lars on May 23, 2015, 11:31:11 am

Quote from: Ttk on May 23, 2015, 11:21:04 am

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

5^5 is 3125 and if each person has one left over then the number has to end in a 1 or 6
I tried it with 3126 but it didnt work so I tried it with 3121 and it did
3121 -1 /5 gave 624. if the first sailor hid it then there would only be 624*4 watermelons left.
repeated this till I got 255 on the fifth time(?). a lot of pain was involved c:

Ah I haven't seen anyone use this approach, very creative.

I personally used a different approach when I first came across this problem. I started from the bottom and figured out that the final amount of coconuts has to be divisible by 5 and 4, in other words a multiple of 20. I also figured out, just like you that every amount of coconuts in the crate has a 1 or a 6 as it's last digit. However I also knew no amount could end with a 1 (except for the starting amount). Thus every upstream amount ends with 6, which means it cannon be a multiple of 40. This is because 40 * 1.25 + 1=51. Which means the final amount of coconuts is a multiple of 20, but not a multiple of 40. I found these limitations sufficient to start using trial and error.

i thought we were discussing watermelons here if it was coconuts it would have been an entirely different thing

Quote from: Lars on May 24, 2015, 02:14:33 pm

Quote from: Ttk on May 23, 2015, 09:57:40 pm

Quote from: Lars on May 23, 2015, 11:31:11 am

Quote from: Ttk on May 23, 2015, 11:21:04 am

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

5^5 is 3125 and if each person has one left over then the number has to end in a 1 or 6
I tried it with 3126 but it didnt work so I tried it with 3121 and it did
3121 -1 /5 gave 624. if the first sailor hid it then there would only be 624*4 watermelons left.
repeated this till I got 255 on the fifth time(?). a lot of pain was involved c:

Ah I haven't seen anyone use this approach, very creative.

I personally used a different approach when I first came across this problem. I started from the bottom and figured out that the final amount of coconuts has to be divisible by 5 and 4, in other words a multiple of 20. I also figured out, just like you that every amount of coconuts in the crate has a 1 or a 6 as it's last digit. However I also knew no amount could end with a 1 (except for the starting amount). Thus every upstream amount ends with 6, which means it cannon be a multiple of 40. This is because 40 * 1.25 + 1=51. Which means the final amount of coconuts is a multiple of 20, but not a multiple of 40. I found these limitations sufficient to start using trial and error.

i thought we were discussing watermelons here if it was coconuts it would have been an entirely different thing

Shit you're right. Forget everything I've said it's all wrong.

Quote from: King125 on May 24, 2015, 02:17:19 pm

Quote from: Lars on May 24, 2015, 02:14:33 pm

Quote from: Ttk on May 23, 2015, 09:57:40 pm

Quote from: Lars on May 23, 2015, 11:31:11 am

Quote from: Ttk on May 23, 2015, 11:21:04 am

The best I can come up with is 3121. If that isn't it I surrender :v

That's the correct answer :p Mind sharing your approach?

5^5 is 3125 and if each person has one left over then the number has to end in a 1 or 6
I tried it with 3126 but it didnt work so I tried it with 3121 and it did
3121 -1 /5 gave 624. if the first sailor hid it then there would only be 624*4 watermelons left.
repeated this till I got 255 on the fifth time(?). a lot of pain was involved c:

Ah I haven't seen anyone use this approach, very creative.

I personally used a different approach when I first came across this problem. I started from the bottom and figured out that the final amount of coconuts has to be divisible by 5 and 4, in other words a multiple of 20. I also figured out, just like you that every amount of coconuts in the crate has a 1 or a 6 as it's last digit. However I also knew no amount could end with a 1 (except for the starting amount). Thus every upstream amount ends with 6, which means it cannon be a multiple of 40. This is because 40 * 1.25 + 1=51. Which means the final amount of coconuts is a multiple of 20, but not a multiple of 40. I found these limitations sufficient to start using trial and error.

i thought we were discussing watermelons here if it was coconuts it would have been an entirely different thing

Shit you're right. Forget everything I've said it's all wrong.

Get your shit together, Lars :kappa: