amistre64 find 2 integers such that they sum to: 1640 and their LCM is: 8400 one year ago one year ago

1. amistre64

my idea is to factor the LCM to create a pool of options

2. amistre64

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

3. CliffSedge

Maybe algebra with a quadratic equation?

4. amistre64

maybe. but number theory methods might be perfered

5. CliffSedge

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.

6. PaxPolaris

$\large x \left( 1640-x \right)=8400n$ where n is a natural number

7. amistre64

hmm, the "n" seems interesting

8. amistre64

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

9. PaxPolaris

$x=820 \pm \sqrt {820^2-8400n}$

10. amistre64

i assume the under radical needs to remain 0 or greater?

11. PaxPolaris

needs to be a perfect square

12. amistre64

lol, yeah i spose that would have to be a major caveat seeing the x needs to be an integer :)

13. PaxPolaris

$\large x=820 \pm 400\sqrt{1681-21n}$

14. amistre64

i think one more condition was such that x and y were both positive values

15. PaxPolaris

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

16. robtobey

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},