Question

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.

Answer 1

\[8400=2^4.3^1.5^2.7^1\]\[1640=2^3~5^1~41^1\]
\[\large a=2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4}\]\[\large b=2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4}\]

\[\large a+b=(2^{m_1}~3^{m_2}~5^{m_3}~7^{m_4})~+~(2^{n_1}~3^{n_2}~5^{n_3}~7^{n_4})=2^3~5^1~41^1\]
\[m_1,n_1\le4\]\[m_2,n_2\le1\]\[m_3,n_3\le2\]\[m_4,n_4\le1\]

divide each side by common factors \(\large 2^3~5^1\)

Answer 2

\[\large \bar a+\bar b=(2^{m_1-3}~3^{m_2}~5^{m_3-1}~7^{m_4})~+~(2^{n_1-3}~3^{n_2}~5^{n_3-1}~7^{n_4})=41\]
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

Answer 3

Oh boy.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

way, waaayyyyy simpler than my trial and error ;)

Answer 11

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