Here's the question you clicked on:
amistre64
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.
\[8400=2^4.3^1.5^2.7^1\]\[1640=2^3~5^1~41^1\] \[\large a=2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4}\]\[\large b=2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4}\] \[\large a+b=(2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4})~+~(2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4})=2^3~5^1~41^1\] \[m_1,n_1\le4\]\[m_2,n_2\le1\]\[m_3,n_3\le2\]\[m_4,n_4\le1\] divide each side by common factors \(\large 2^3~5^1\)
\[\large \bar a+\bar b=(2^{m_1-3}~3^{m_2}~5^{m_3-1}~7^{m_4})~+~(2^{n_1-3}~3^{n_2}~5^{n_3-1}~7^{n_4})=41\] 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
im not usually that impressed in math class, but i simply feel in love with this :)
It's surprising to hear that THE AMISTRE64 has a teacher!
lol, well it is college afterall. we cant just teach ourselves ;)
W-w-w-wait... dad in college?
\[\begin{matrix} 2&3&5&7&\bar a& \bar b\\ -&-&-&-&-&-\\ 0&0&0&0&1&210&\ne41\\ 0&0&0&1&7&30&\ne41\\ 0&0&1&0&5&42&\ne41\\ 0&0&1&1&35&6&=41&found\ it\\ 0&1&0&0\\ 0&1&0&1\\ 0&1&1&0\\ 0&1&1&1\\ \end{matrix}\]
\[a=\bar a~2^3~5=35(40)=1400\]\[b=\bar b~2^3~5=6(40)=240\]
way, waaayyyyy simpler than my trial and error ;)
im 5 years (fingers crossed) away from a masters in math