Help with algebra question
[From: ] [author: ] [Date: 13-01-21] [Hit: ]
Then we have.. c = ((38m+2)(38n+4) - 8)/38 = 38m*n + 2n + 4mThis gives us a set of solutions that include.. (a, b,......
x=42 and not -2 since only even positive integers
therefore the answer is 40;42;44.
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You seem to require only a, b, c such that
.. a*b = 38c + 8 ... where a, b, c are positive even integers. (There is no requirement for a, b, c to have any particular relationship to each other.)
This gives
.. c = (ab - 8)/38
We need to have "a" and "b" so that a*b mod 38 = 8. We can choose a=38m+2 and b=38n+4 where "2" and "4" are even factors of 8, and m and n are any positive integers. Then we have
.. c = ((38m+2)(38n+4) - 8)/38 = 38m*n + 2n + 4m
This gives us a set of solutions that include
.. (a, b, c) = (2, 42, 2), (40, 4, 4), (40, 42, 44), (40, 80, 84) ... <== ANSWERS
note that 40, 42, 44 is a set of consecutive even integers satisfying the requirements.
We can also choose a=38m+6, b=38n+14. Then we have
.. c = 38m*n + 14m + 6n + 2
This gives rise to another set of solutions, including
.. (a, b, c) = (44, 14, 16), (6, 52, 8), (44, 52, 60), (44, 90, 104), ... <== ANSWERS
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In this last formulation, we have 6*14 mod 38 = 8. It turns out that 8 added to any even multiple of 38 will have at least one set of factors "j" and "k" so that we can formulate solutions to your problem in the form
.. a=38m+2j, b=38n+2k
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Sounds like you need to put an equation together. Off of the top of my head, it seems you would use the equation:
a * b = (38 * c) - 8
If you plug in 2 of the variables, you should be able to solve for the third.
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x
x + 2
x + 4
x(x + 2) = 38(x + 4) + 8
x^2 + 2x = 38x + 152 + 8
x^2 - 36x - 160 = 0
(x - 40)(x + 4) = 0
x = -4, 40
ignore negative solution
x = 40
40, 42, 44
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A,B,C are your three integers.
AB=38C - 8
This can be manipulated to get an answer for each:
A = (38C - 8) / B
B = (38C - 8) / A
C = (AB + 8) / 38
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x-2 , x , x+2
solve for x in x(x-2) = 38(x+2)+8
