anonymous
  • anonymous
Fool's problem of the day, If \( m \) and \(n\) are positive integers that satisfy \(3m^2 − 8n^2 + 3m^2n^2 = 2008\), then find the product \(m^2n\). Genre: algebra-precalculus Rating: Easy
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
asnaseer
  • asnaseer
I have a feeling that we should make use of the fact that \(2008=2^3*251\)
asnaseer
  • asnaseer
I don't know if this helps, but we can rewrite this as:\[3m^2(1+n^2)-2^3n^2=2^3*251\]
asnaseer
  • asnaseer
therefore:\[3m^2(1+n^2)=2^3(251+n^2)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Good point asnaseer.
asnaseer
  • asnaseer
therefore:\[m^2n=\frac{2^3n(251+n^2)}{3(1+n^2)}\]
asnaseer
  • asnaseer
this must be an integer, therefore \(3(1+n^2)\) must divide evenly into \(2^3n(251+n^2)\)
anonymous
  • anonymous
You are in the right track, divisibility the way :)
asnaseer
  • asnaseer
since \(2^3\) is not divisible by 3, then this implies \(n(251+n^2)\) must be divisible by 3. therefore either n is divisible by 3 and/or \(251+n^2\) is divisible by 3.
anonymous
  • anonymous
is it 112? (n = 7), makes (251+n^2) =300 which is nice and makes (1+n^2)=50 also nice :D
asnaseer
  • asnaseer
I'm trying to /prove/ the result. I am sure we can use brute force to find /a/ solution, but that doesn't prove it's the /only/ solution. :)
anonymous
  • anonymous
ah, nvm them, I don't know how to prove it :D
anonymous
  • anonymous
Yes 112 is the answer :)
asnaseer
  • asnaseer
you win gogind :D
anonymous
  • anonymous
win by brute force, sounds like cheating :D
anonymous
  • anonymous
haha, prove it now :D
asnaseer
  • asnaseer
need fooood - I'll be back soon :)
anonymous
  • anonymous
sure :)
anonymous
  • anonymous
Good thinking gogind.
asnaseer
  • asnaseer
OK - the furthest I can get with this is as follows:\[\begin{align} 3m^2(1+n^2)-8n^2&=2008\tag{a}\\ m^2n&=\frac{8n(251+n^2)}{3(1+n^2)}\tag{b} \end{align}\]equation (b) implies \(n(251+n^2)\) must be a multiple of 3. 1. lets try n being a multiple of 3.\[\begin{align} n&=3a\\ 3m^2(1+9a^2)-72a^2&=2008\qquad\text{(substituting n=3a into equation(a))}\\ \therefore 3(m^2(1+9a^2)-24a^2)&=2008\\ \qquad&\text{reject since 2008 is not a multiple of 3} \end{align}\] 2. lets try \(251+n^2\) being a multiple of 3.\[\begin{align} 251+n^2&=3a\\ \therefore n^2&=3a-251\tag{c} \end{align}\] Now we can substitute equation (c) into equation (a) to get:\[\begin{align} 3m^2(1+3a-251)-8(3a-251)&=2008\\ 3m^2(3a-250)-24a+\cancel{2008}&=\cancel{2008}\\ 3m^2(3a-250)&=24a\\ \therefore m^2&=\frac{8a}{3a-250}\tag{d} \end{align}\] Equation (d) implies \(a\ge84\). It also shows \(m^2\rightarrow\frac{8}{3}\) as \(a\rightarrow\infty\). From here I could only use trial and error to determine that the only solution to (c) and (d) is a=100, which gives m=4, n=7. Therefore \(m^2n=112\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.