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14th January 2008, 07:46 PM #1GOLD MEMBER
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See if you're going to seed....another '7' problem!
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???
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14th January 2008, 08:49 PM #2You've got to risk it to get the biscuit
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14th January 2008, 08:49 PM #3
7? :d
Chris
========================================
Life isn't always fair
....................but it's better than the alternative.
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14th January 2008, 09:01 PM #4You've got to risk it to get the biscuit
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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???S T I R L O
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14th January 2008, 09:49 PM #5
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?
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14th January 2008, 09:54 PM #6
He either planted 6 or 1 or 0.
The reason is that the problem generates a quadratic equation. Which generates three possible answers.Have a nice day - Cheers
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14th January 2008, 10:22 PM #7You've got to risk it to get the biscuit
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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 todayS T I R L O
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14th January 2008, 10:27 PM #8
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 .Have a nice day - Cheers
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14th January 2008, 10:32 PM #9You've got to risk it to get the biscuit
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14th January 2008, 11:07 PM #10You've got to risk it to get the biscuit
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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 ...S T I R L O
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14th January 2008, 11:12 PM #11
Alt codes for powers.
(Hold the alt key down & use the numbers on the keypad)
Atl 0178 ²
Alt 0179 ³Cliff.
If you find a post of mine that is missing a pic that you'd like to see, let me know & I'll see if I can find a copy.
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14th January 2008, 11:14 PM #12
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.Have a nice day - Cheers
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14th January 2008, 11:17 PM #13You've got to risk it to get the biscuit
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²³
ta cliff.S T I R L O
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14th January 2008, 11:22 PM #14You've got to risk it to get the biscuit
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15th January 2008, 07:42 AM #15GOLD MEMBER
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'6' and you're in
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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