See if you're going to seed....another '7' problem!

Quote:

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Originally Posted by Greg Ward

6 is the answer, for the gardener planted seeds (plural).

Any cubed number divided by 7 has only 3 remainders (not reminders Wongo, but I shared your excitement yesterday): 0, 1 or 6

1 cubed: 1/7. remainder; 1
2 cubed: 8 /7 remainder: 1
3 cubed: 27/7 remainder: 6
4 cubed: 64/7. remainder: 1
5 cubed: 125/7 remainder: 6
6 cubed: 216/7 remainder: 6
7 cubed: 343/7 remainder: 0

Can anyone prove that these are the only 3 solutions for all cubed numbers divided by 7?


Greg

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The Butcher got his cuts mixed up, but basically already gave you the answer: the problem is a cubic function that has three real roots (you can check on wikipedia, this time they got it right:D).

Quote:

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Originally Posted by Greg Ward

Any cubed number divided by 7 has only 3 remainders (not reminders Wongo, but I shared your excitement yesterday): 0, 1 or 6

1 cubed: 1/7. remainder; 1
2 cubed: 8 /7 remainder: 1
3 cubed: 27/7 remainder: 6
4 cubed: 64/7. remainder: 1
5 cubed: 125/7 remainder: 6
6 cubed: 216/7 remainder: 6
7 cubed: 343/7 remainder: 0

Can anyone prove that these are the only 3 solutions for all cubed numbers divided by 7?

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14 cubed: 2744 / 7 remainder: 0
21 cubed: 9261 / 7 remainder: 0
28 cubed: 21952 / 7 remainder: 0
35 cubed: 42875 / 7 remainder: 0

I found 4.:U

Quote:

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Originally Posted by Wongo

14 cubed: 2744 / 7 remainder: 0
21 cubed: 9261 / 7 remainder: 0
28 cubed: 21952 / 7 remainder: 0
35 cubed: 42875 / 7 remainder: 0

I found 4.:U

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:?

Can anyone prove that these are the only 3 solutions for all cubed numbers divided by 7?

NOT

Can anyone prove that there are the only 3 solutions for all cubed numbers divided by 7?


Sorry I read the question incorrectly. I knew it was something wrong with it.:doh:

See I am not perfect after all. :D

Such good reputation huh.

My bad.