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| A Pythagorean triplet (Posted on 2012-08-28) |
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Given that a2+b2 = c2
find the value of a*b*c if a+b+c=1000.
Pen & paper approach (spoiler)
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Comment 4 of 4 | 
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I was intrigued that the previous computer posts could find only two solutions for positive values of a, b and c, so took to pen and paper:
(a + b)2 = a2 + b2 + 2ab = c2 + 2ab (1)
(1000 – a)(1000 – b) = (b + c)(c + a)
= c(a + b + c) + ab
= 1000c + ab
= 1000c + [(a + b)2 – c2]/2 using (1)
= 1000c + [(1000 – c)2 – c2]/2
(1000 – a)(1000 – b) = 500000 (2)
500000 = (25)(56), so its factors must all be of the form 2x5y where x and y
range from 0 to 5 and 6 respectively, giving 6*7 = 42 ordered pairs as follows.
(1,500000),....,(500,1000), (625,800), (800,625),....,(1000,500),(500000,1).
There are also 42 corresponding pairs of negative factors, giving a total of 84 solutions to equation (2).
If the two brackets in (2) are positive then a and b must both be less than 1000, so the only factors possible are the underlined central ones, giving
1000 – a = 625 and 1000 – b = 800 from which (a,b,c) = (375,200,425)
or 1000 – a = 800 and 1000 – b = 625 from which (a,b,c) = (200,375,425).
The other 82 possible pairs give solutions involving negative values for one of a, b and c, as listed in previous posts.
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Posted by Harry
on 2012-08-30 19:18:15 |
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