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Topic: Product and Sum Triplet Product Difference Puzzle (Read 400 times) |
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K Sengupta
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Product and Sum Triplet Product Difference Puzzle
« on: Dec 24th, 2009, 8:16am » |
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Determine all possible quadruplet(s) (a, b, c, d) of positive integers with a <=b <= c <= d, that satisfy this equation:
a*b*c*d – (a*b*c + a*b*d+ b*c*d +a*c*d) = 110
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| « Last Edit: Dec 24th, 2009, 10:16am by K Sengupta » |
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R
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Re: Product and Sum Triplet Product Difference Puz
« Reply #1 on: Dec 24th, 2009, 8:35am » |
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But why
Did you forget something, like a,b,c and d are digits not integers? And abcd means a*b*c*d, if they are integers, not digits.
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ThudnBlunder
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Re: Product and Sum Triplet Product Difference Puz
« Reply #2 on: Dec 24th, 2009, 8:56am » |
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on Dec 24th, 2009, 8:35am, R wrote:
Did you forget something, like a,b,c and d are digits not integers? |
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I think he intends them to be integers.
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K Sengupta
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Re: Product and Sum Triplet Product Difference Puz
« Reply #3 on: Dec 24th, 2009, 10:18am » |
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on Dec 24th, 2009, 8:35am, R wrote:
abcd means a*b*c*d, if they are integers, not digits. |
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I confirm having suitably amended the problem text to obviate the ambiguity.
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| « Last Edit: Dec 24th, 2009, 10:18am by K Sengupta » |
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R
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Re: Product and Sum Triplet Product Difference Puz
« Reply #4 on: Dec 24th, 2009, 1:26pm » |
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If a,b,c and d are the roots of the quartic equation P0x4 + P1x3 + P2x2 + P3x1 + P4 = 0, then the given condition tells us that:
P3 + P4 = 110 P0
That's all I could get. There seems to be like many such quadruplets.
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MathsForFun
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Re: Product and Sum Triplet Product Difference Puz
« Reply #5 on: Dec 27th, 2009, 1:57am » |
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I would guess that the solution lies in limits to what the divisors of a, b, c and d must be. Not sure if this is any help, but:
a = b*c*d/(((b-1)*c-b)*d-b*c)
b = a*c*d/(((a-1)*c-a)*d-a*c)
c = a*b*d/(((a-1)*b-a)*d-a*b)
d = a*b*c/(((a-1)*b-a)*c-a*b)
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aicoped
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Re: Product and Sum Triplet Product Difference Puz
« Reply #6 on: Jan 25th, 2010, 1:14pm » |
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I have no clue how to solve this, but trial and error I got 4,5,5,6 as a solution.
EDIT
also
2,4,5,75
2,5,5,32
EDIT number 2
Ok my thoughts.
a must be bigger than 1, since you will always return a negative number with the other three answers being postive. A can't be higher than 4, since 5,5,5,5 is 125 and any other numbers will all be bigger than that. So my range is now limited to:
2<=a<=4
Once you have that, you can find maximum values for b and then test all values beneath that and repeat for c.
For example if a=2, then b can not be bigger than 7, for 2,8,8,8 yields a number higher than 110. So i place it equal to 2 find that my results will always be negative, so i set to 3 and find the same, at 4 I am finally able to get a positie answer,and then find out the maximal c value and so on.
This is very cumbersome, but I can't think of a better way.
looking at the field of answers so far:
2,4,5,75
2,5,5,32
4,5,5,6
It would seem if there are any more ansers they would have to be in the 3,x,x,x range or possibly higher than 2,5,5,x. I think I did all of them by hand but i might have missed some.
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| « Last Edit: Jan 25th, 2010, 9:27pm by aicoped » |
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