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

[Greg Ward]

A gardener collecting seed pods found a many pods on each plant as there was plants

Each pod had in it as many seeds as there was plants

The gardener shared the seeds between seven friends, giving each the same number and as many as possible.

The seeds which remained he planted in his own garden.

How many seeds did he plant???

[ss_11000]

A gardener collecting seed pods found as many pods on each plant as there was plants

Each pod had in it as many seeds as there was plants

The gardener shared the seeds between seven friends, giving each the same number and as many as possible.

The seeds which remained he planted in his own garden.

How many seeds did he plant???
zero?:?

eg
7 plants = 49 pods = 343 seeds
each friend gets 49 seeds and the farmer gets none.

but the question says he planted the remaining seeds implying that the answer cannot be zero or one. so i'm wrong.:(

[chrisb691]

7? :d

[ss_11000]

plants = x

# of pods = x^2
# of seeds = x^2 * x^2

7 friends means the number of seed each is given is:

( 2x^2 )
______
``7

am i right so far?

if so, then we dont have enough info to give a number of seeds of which the farmer planted. do we???:?

[AlexS]

plants = x

``7

am i right so far?



No
Let no. of plants = x
then No. of seeds =x^3

If x=2
x^3=8
therefore, the farmer is left with 1 seed

but, if x=3
x^3 =27, and the farmer is left with 6 seeds

and if x = 7, he will be left with no seeds.

Try a few other numbers. Is he ever left with anything other than 1, 6 or 0? Why?

[Wood Butcher]

He either planted 6 or 1 or 0.

The reason is that the problem generates a quadratic equation. Which generates three possible answers.

[ss_11000]

oh, you guys are smart!

i came first in maths ( yr 10 ) at my school so that goes to show you how smart kids are today:D

[Wood Butcher]

Stirlo, given that the algebra to justify the answer is Yr 12 Advanced Maths level don't feel to hard on yourself ;).

And don't ask me too. I just know a lot of shortcuts to use in spreadsheets :D.

[ss_11000]

Stirlo, given that the algebra to justify the answer is Yr 12 Advanced Maths level don't feel to hard on yourself ;).

And don't ask me too. I just know a lot of shortcuts to use in spreadsheets :D.

well i'm doing 3 unit maths this year so ask me again nexty year and i will be able to tell you how to do it without spreadsheets.:cool:

but do you know whats cool. i understand what you and Alex are saying:2tsup:

[ss_11000]

http://www.abc.net.au/canberra/stories/s741130.htm

link to the same question.

they rekon the answer has to be six.

but the formula is buggin me. i cant work it out:( how do you write a formula saying that the remainder is the answer - or am i looking at it wrong.

eg (x^3) / 7 ...

[Cliff Rogers]

Alt codes for powers.

(Hold the alt key down & use the numbers on the keypad)

Atl 0178 ²
Alt 0179 ³

[Wood Butcher]

I think it would be something like this:

z = x³ - 7y

where:
z is the no of seeds he planted
x is the number of plants
y is the number of seeds given to his friends

The ABC site is a little off saying that the correct answer is 6 because 1 is also correct.

[ss_11000]

²³
ta cliff.

[ss_11000]

I think it would be something like this:

z = x³ - 7y

where:
z is the no of seeds he planted
x is the number of plants
y is the number of seeds given to his friends
okay, that formula works.:2tsup:



The ABC site is a little off saying that the correct answer is 6 because 1 is also correct.

yeah, and they cant get off on a technicality besuase they dont use the word "seeds" as a plural to not accept one being correct.

[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

[Frank&Earnest]

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

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).

[Wongo]

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?


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

[Frank&Earnest]

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

:?

[Wongo]

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.