## amistre64 4 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.

1. amistre64

$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

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

Oh boy.

4. amistre64

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

5. ParthKohli

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

6. amistre64

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

7. ParthKohli

8. amistre64

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

9. amistre64

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

10. amistre64

way, waaayyyyy simpler than my trial and error ;)

11. amistre64

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