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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???
zero?:?Quote:
Originally Posted by Greg Ward
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.:(
7? :d
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???:?
NoQuote:
Originally Posted by ss_11000
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?
He either planted 6 or 1 or 0.
The reason is that the problem generates a quadratic equation. Which generates three possible answers.
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
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:Quote:
Originally Posted by Wood Butcher
but do you know whats cool. i understand what you and Alex are saying:2tsup:
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 ...
Alt codes for powers.
(Hold the alt key down & use the numbers on the keypad)
Atl 0178 ²
Alt 0179 ³
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.
²³
ta cliff.
okay, that formula works.:2tsup:Quote:
Originally Posted by Wood Butcher
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.Quote:
The ABC site is a little off saying that the correct answer is 6 because 1 is also correct.
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).Quote:
Originally Posted by Greg Ward
14 cubed: 2744 / 7 remainder: 0Quote:
Originally Posted by Greg Ward
21 cubed: 9261 / 7 remainder: 0
28 cubed: 21952 / 7 remainder: 0
35 cubed: 42875 / 7 remainder: 0
I found 4.:U
:?Quote:
Originally Posted by 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.