Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

amistre64

  • 2 years ago

find 2 positive intergers, "a" and "b", such that: a+b = 1640 LCM[a,b] = 8400 my trial and error method got a right answer, but the teacher showed a pretty cool way to get it.

  • This Question is Closed
  1. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[8400=2^4.3^1.5^2.7^1\]\[1640=2^3~5^1~41^1\] \[\large a=2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4}\]\[\large b=2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4}\] \[\large a+b=(2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4})~+~(2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4})=2^3~5^1~41^1\] \[m_1,n_1\le4\]\[m_2,n_2\le1\]\[m_3,n_3\le2\]\[m_4,n_4\le1\] divide each side by common factors \(\large 2^3~5^1\)

  2. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\large \bar a+\bar b=(2^{m_1-3}~3^{m_2}~5^{m_3-1}~7^{m_4})~+~(2^{n_1-3}~3^{n_2}~5^{n_3-1}~7^{n_4})=41\] all exponents then reduce to possibilities of 0 or 1, and there are no common factors between a and b at this point therefore we can systematically go thru and find a solution

  3. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh boy.

  4. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    im not usually that impressed in math class, but i simply feel in love with this :)

  5. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's surprising to hear that THE AMISTRE64 has a teacher!

  6. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    lol, well it is college afterall. we cant just teach ourselves ;)

  7. ParthKohli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    W-w-w-wait... dad in college?

  8. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[\begin{matrix} 2&3&5&7&\bar a& \bar b\\ -&-&-&-&-&-\\ 0&0&0&0&1&210&\ne41\\ 0&0&0&1&7&30&\ne41\\ 0&0&1&0&5&42&\ne41\\ 0&0&1&1&35&6&=41&found\ it\\ 0&1&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&1&1&1\\ \end{matrix}\]

  9. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \[a=\bar a~2^3~5=35(40)=1400\]\[b=\bar b~2^3~5=6(40)=240\]

  10. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    way, waaayyyyy simpler than my trial and error ;)

  11. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    im 5 years (fingers crossed) away from a masters in math

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.