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find 2 integers such that they sum to: 1640 and their LCM is: 8400

Mathematics
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my idea is to factor the LCM to create a pool of options
8400 = 84*100 = 84*2*2*5*5 = 2*2*3*7*2*2*5*5 2*2*2*2*3*5*5*7, therefore 2 2*2 2*2*2 2*2*2*2 2*2*2*2*3 2*2*2*2*3*5 2*2*2*2*3*5*5 2*2*2*2*3*5*5*7 etc ... but that does seem like a long way around it
Maybe algebra with a quadratic equation?

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Other answers:

maybe. but number theory methods might be perfered
Might want to use optimization methods from calculus too to minimize the coefficients. Yes, it does seem like a number theory issue, but algebra is always my starting point for solving unknowns.
\[\large x \left( 1640-x \right)=8400n\] where n is a natural number
hmm, the "n" seems interesting
on the test i just ended up using the y= on my ti83 y1 = 1640-x y2= lcm(x,1640-x) then searched thru that table till i found y2 = 8400
\[x=820 \pm \sqrt {820^2-8400n}\]
i assume the under radical needs to remain 0 or greater?
needs to be a perfect square
lol, yeah i spose that would have to be a major caveat seeing the x needs to be an integer :)
\[\large x=820 \pm 400\sqrt{1681-21n}\]
i think one more condition was such that x and y were both positive values
sorry, \[\large x=820 \pm \sqrt{400(1681-21n)}\]\[\large x=820 \pm 20\sqrt{1681-21n}\] and n is the GCF of x and y
240 + 1400 = 1640, LCM[240, 1400] = 8400 Used the following small program and fed it to Mathematica. Table[{n, LCM[n, 1640 - n] == 8400}, {n, 820}] Part of the Output: {233, False}, {234, False}, {235, False}, {236, False}, {237, False}, \ {238, False}, {239, False}, {240, True}, {241, False}, {242, False}, \ {243, False}, {244, False},

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