402 0x700
With a little bit of basic algebra, the terms can be moved around for each
line so the remainder (shown in bold) is by itself on the left of the equal sign:53 = 7253 −60 · 120
14 = 120− 2·53
11 =53− 3·14
3 =14−1·11
2 =11−3·3
1 =3− 1·2
Starting from the bottom, it’s clear that:
1=3−1· 2
The line above that, though, is 2 = 11 − 3 · 3, which gives a substitution
for 2:1=3−1·(11− 3·3)
1=4· 3 − 1·11
The line above that shows that 3 = 14 − 1 · 11, which can also be
substituted in for 3:1=4·(14− 1·11)−1·11
1=4·14−5· 11
Of course, the line above that shows that 11 = 53 − 3 · 14, prompting
another substitution:1=4·14−5·(53− 3·14)
1=19· 14 −5·53
Following the pattern, we use the line that shows 14 = 120 − 2 · 53,
resulting in another substitution:1=19·(120− 2 · 53)−5·53
1=19·120− 43 · 53
And finally, the top line shows that 53 = 7253 − 60 · 120, for a final
substitution:1=19·120− 43 · (7253− 60 · 120)
1 = 2599 · 120− 43 · 7253
2599 · 120 + −43 · 7253 = 1
This shows that J and K would be 2599 and −43, respectively.