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amistre64
Group Title
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.
 one year ago
 one year ago
amistre64 Group Title
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.
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.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 \(\large 2^3~5^1\)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
\[\large \bar a+\bar b=(2^{m_13}~3^{m_2}~5^{m_31}~7^{m_4})~+~(2^{n_13}~3^{n_2}~5^{n_31}~7^{n_4})=41\] 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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh boy.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
im not usually that impressed in math class, but i simply feel in love with this :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
It's surprising to hear that THE AMISTRE64 has a teacher!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
lol, well it is college afterall. we cant just teach ourselves ;)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Wwwwait... dad in college?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
\[\begin{matrix} 2&3&5&7&\bar a& \bar b\\ &&&&&\\ 0&0&0&0&1&210&\ne41\\ 0&0&0&1&7&30&\ne41\\ 0&0&1&0&5&42&\ne41\\ 0&0&1&1&35&6&=41&found\ it\\ 0&1&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&1&1&1\\ \end{matrix}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
\[a=\bar a~2^3~5=35(40)=1400\]\[b=\bar b~2^3~5=6(40)=240\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
way, waaayyyyy simpler than my trial and error ;)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
im 5 years (fingers crossed) away from a masters in math
 one year ago
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